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hi)
HARVARD COLLEGE
SCIENCE CENTER
LIBRARY
ELEMENTS
OF
ALGEBRA.
MAJOR D. H. HILL,
tMfonB or MtTBniitia in am umuiiii iHg n lUTiMim vxiagt, a. a;
un nwriMm or >uHnuxm vt wamkomb ooiun^ t4.
/vVt/
PHILADELPHIA:
J. B. LIPPINCOTT & CO.
1857.
IVatJsforred f'rom
The Iiswvonoe Bcientino BchooU
O Febrnnryv 1B8B.
Entered, aooording to the Act of Congrew, in the year 1857,^ by
J. B. LIPPINCOTT St CO.,
In the Clerk's Office of the DiBtriot Court of the United SUtei for the EMtem
BiBtriet of PennaylTania.
muoimD n j. vieur.
TESTIMONIALS.
jFVom T. J. Jackaonj Pro/esior of NaJtwraL and JSxperimenUU F%%Uh
iophy^ Virginia Military InsHhUe.
"Fnm. an examination of yarionB portionB of M%Jor B. H. Hill's Algebra^ in
mannaeript, I regard it as superior to any other work with whioh I am
aoqnainted on the same branch of soienoe."
Ik-om a Teacher of MaAemaHcs.
** Haying also examined seyeral chapters of Mfljor Hill's Algebra^ I hare no
heritation in oononrring in the aboTC opinion of Professor Jackson."
WILLIAM Mclaughlin.
From J, L. Campbdlf Professor of Natural Science^ WaAington
CoQege^ Virginia.
" While I ftdly concur with Professor Jackson and Mr. McLaughlin in the
opinion they express of Major Hill's Algebra, I will add, that I regard the
method adopted by the author, of incoiporating into the work some of the
elementary principles of the Calculus, as giying it peculiar yalue as a college
text-book."
From William Oxlhamj Professor of Chemistry and Oeohgyj Tirginia
Military Institute, and late Profemor at West Point, N". T,
** Haying read the greater portion of Major Hill's Algebra, I consider it as
better adapted for use as a college text-book than any work on the subject With
which I am acquainted."
From Professor J. A. Lektnd, late Professor of MoUhematies in the
State Military Institute of S. C.
« I haye examined with care many of the proof sheets of Mfljor Hill's Algebra,
and it affords me pleasure to concur in the fayorable opinions aboye expressed."
ir TESTIMONIALS.
This work of Professor HiU's is the product of a mind intensely in love iritli
Algebra. It bears the marks of unremitting and intelligent toiL It is ezhans-
tire on the subjects it treats ; and, in the abundance and aptness of its illustra-
tioDS, reminds one of the richness and simplicity of Euler.
GSARLIB PHILLIPS,
Prof. CM EngiMeHng^ Unwrn-rily^ N. C.
CONTENTS
I>EPINinONS AND PREUMINABT RBlfARKS 18
ADDITION 18
Can L When the Terms are like, and haye like Signs • 18
IL When the Quantities are like, but affeoted with unlike Signs.. 19
in. When the Qoantities are similar and dissiniilar, and haye like
and nnlike Signs ..,.•...•«.• -...••......-• 20
0UBTBACTION ^ .• 21
MULTIPLICATION „ 24
Com L When both Mnltiplieand and Multiplier are Simple Quantities 2i
n. When the Moltiplicand is a Compound Quantity, and the
Multiplier a Monomial 27
nL When the Multiplicand and MuUiplier are both Compound
Quantities.. ...... .••...... ^..•••. 29
Thtorenu ^ 88
Faetoritiff PolywmiaU •• ^. • ••• 86
MVISION . „.• 88
CoH I. When the Diyidend and Diyisor are both Monomials ..-..,.•... 88
n. When the Diyidend is a Compound Quantity, and the Diyisor
a Monomial « • ^ 41
m. When the Diyidend and Diyisor are both Po^omials ....... .^ 42
Prine^Ui m DwitUm ^ ........ 48
AIaJAIS&AIv fAACTIONSa.***. •••••••••••••••••••.••••••• •••»mmmm«m ..«••••••«•••«• ol
AVUSIillVW Wf JFrW!mff99 .. .•.••«..• •••.••.•• #•••*...• •.•••..«• •••••«•.• ••••••••• •••.•...• ••• 04
Com L To reduce a ffimple Fraction to its Simplest Form 64
n. To reduce a Mixed Quantity to the Form of a Fraction. 66
IIL To reduce a Fraction to an Entii« or Bfized Quantity 66
!♦ (y)
Tl CONTENTS.
ISeAidibfi^/WMliMM (oontinued).
Com IV. To derelop a FraotioA into a Series 67
y. To reduce Fractions haTing Different Denominators to Equi-
valent Fractions having the same Denominator . »• 69
YI. To add Fractional Quantities together 61
yiL To Snbtract one or more Fractional Quantities from one or
more Fractional Quantities ..« 62
ym. Jo multiply Fractional Quantities together 64
IX. To dlYide Fractional Quantities bj each other .«...•..• 65
X. To reduce a Compound Fraction to its Lowest Tenns .• 68
Leoit Common MvUiple,,,, .• .• 76
Corollary ; 78
LeaH Common MvUipU oflVaetiont. 79
CfreaUti Common Dmtor of Draetiom , • 80
EQUATIONS OF THE FIRST DEGREE 82
SotuUon ofSquaiioni of the FirH Degree 86
First Transformation — clearing an Equation of Fractions 86
Second Transformation — transposing Terms 86
Third Transformation — clearing the Unknown Quantity of Coefficients 86
Solutions of Problems producing Equations of the First Degree with aa
Unknown Quantity 91
GEOMETRICAL PROPORTION 95
Theorems •« ••... 96
General Remarks - 108
Problems in Geometrical Proportion - 105
NEGATIVE QUANTITIES , 107
Problem of the Couriers • 110
General Problems 116
ELIMINATION BETWEEN SIMULTANEOUS EQUATIONS OF THE
FIRST DEGREE 129
Elimination by Comparison 180
Elimination by Addition and Subtraction 181
Elimination by the Greatest Common Dirisor 188
Examples in Elimination between two Simple Equations of the First
Degree inyolring two Unknown Quantities 185
Elimination between any Number of Simultaneous Equations 140
General Remarks. 149
Problems producing Simultaneous Equations of the First Degree 148
00NTIVT8. nt
YANISHINa FRACTIONS ^ 169
FORMATION OF THE POWERS AND EXTRACTION OF ROOTS 168
Of IneoBuiwiisuable NmnlMrs 176
Extraction of the Square Root of FraotioiiB ^ 177
Mixed Nninbers...... 184
Roots of Numbers entirely Decimal. ^ 186
Square Root of Fractions expressed deoimnlly -• 187
Extraction of the Cabe Root of Numbers ^ 188
Approximate Roots of Incommensurable Numbers 196
Cube Root of Fractions ^ 197
Approximate Root to within a certain Decimal *• 199
Cam L Approximate Root of Whole Numbers to nithin a certain
Decimal ^ - 199
n. Approximate Roots of Mixed Numbers to irithin a cer-
tain Decimal M - 200
Approximate Root of DecinuJ Fractions to within a certain Decimal... 202
Approximate Root of Vulgar Fractions to within a certun Decimal 208
General Remarks on the Extraction of the Cube Root ••... 204
Extraction ^ftht Square Boot of Algebraic Quantitiee m- 206
Square Root of Monomials ^ ^ 206
Square Root of Polynomials. ^ 207
Square Root of a Polynomial inTolving Negatire Exponents 214
Square Root of Incommensurable Polynomials 216
Square Root of Polynomials containing Terms affected with Fractional
Exponents .• 217
Cube Boot of Polynomiale ^ ^ 217
Cube Root of Incommensurable Polynomials 222
Cube Root of Polynomials inyoMng Fractional Exponents. .•••••••• 228
Cube Root of Polynomials containing one or more Terms affected with
NegatiTc Exponents. ••.... 224
Quantitiee afeeted with I^aetional Exponmte .« 226
Multiplication of Quantities affected with Entire or Fractional Ex-
ponents 227
DiTision of Quantities affected with Fractional and Entire Exponents —
Monomials. •«•... 228
Basing to Powers Quantities affected with Fractional and Entire Ex-
ponents — Monomials • .,.t 280
imi CONTXV78.
ilffeeUd wUh DraetunuU JExponmtt (oontintnd).
Extraction of Boots of QnantitieB affeoted ivith Entire or Fractional
EzponentB — Monomials. ^ ^ 282
Promiacnons Examplea^— • ..- -m. •—•••«• •••••• •••««• ••••«•«•• 288
CaUuluB of Badiedli ^ ^ 286
First Principle ^ 288
Second Principle 240
Third Principle ; 241
Fonrth Principle *.' 242
FSffli Principle ....^ 244
ffiztii Principal ^ ^ 247
Seventli Principle. 248
Eighth Principle ^ 260
Tv flMffS Sutdi fiUiondl bp MuU^UcttUon •••••• ••••••••••••«•••••••••••••• 253
{7(SM L Monomial Surds*. ••••••••••••••••••••••••••••^••••••••••••••••••••••••••* 268
Corollaxy. « ^..^••••m 264
n. To find a Multiplier that iriU make rational an E^qwessioft
consisting of a Monomial Snrd connected with Bational
Terms, or consisting of two 'Monomial Snrds 266
Corollary 261
in. To make rational an Expression containing three or mora
Terms of the Square Boot —••.•- .^.m 262
JExiraetitm of the Square Soot of a Monomial Surd eonneeted toith a SaHoiuU
Term, or qf two Monomial Surdt, - 266
First Principle -....•...> 266
Second Principle ••••••••••• ••«.••••. •••••••••••• .«.••••• ••••• • 266
Third Principle ^ - •«..... 266
Inuiffinarjf (iuanHHi9 270
■QUATIONS OF THE SECOND DEGBEE 276
Incomplete Equations —•••••••- 276
Properties of Incomplete Equations « 284
Binomial Equations - • ^ - 286
General Problems in Binomial Equations ••••••••• ••• •••«•• •••••« ••••••••• 287
Complete Equations of the Second Degree.. •••...-« -•.••••-.•-.....• 289
FirH Property, Erety Complete Equation of the Second Degree has
two Values, and but two, for the Unknown Quantity ••»• 296
OONTSNT0. IX
EQUATIONS OF THE SECOND DEGREE —
Complete Eqiutioiis of the Second Degree (eofitmned).
Bteotid Property. The First Member of every Equation of the Second
Degree can be decomposed into two Factors of the First Degree
with respect to ar, the First Factor being the Algebraic Sum of x
and the First Value of x with its Sign changed; and the Second
Factor being the Algebraic Sam of x and the Second Yalne with
its Sign changed. The Sec<md Member of the Equation after this
Decomposition will be Zero....... » .• 296
Tkird Frpptriy. The Algebraic Sum of the VahieB of eTery Com*
plete Equation of the Second Degree is equal to the Coe£Eicient of
the First Power of the Unknown Quantity, with its Sign changed.. 298
Fourth Property, The Product of the Values in every Complete
Equation of the Form, x^ -^ jkb = ^, is equal to the Second Mem-
ber or Absolute Term, with its Sign changed 298
Pfflh Property. The Value of x, in every Complete Equation of the
Second Degree, is half the Coefficient of the First Power of
Xj plus or minus the Square Boot of the Square of half this Co^
efficient increased by the Absolute Term ••••••.... 299
Ckneral Examples ...., ».».••••• •••t.*.. •... 800
Discussion of Complete Equations of the Second Degree • 804
Lralional, Imaginaiy, and Equal Values. • 807
Suppositions made upon the Constants 808
Explanation of Imaginary Values. .• • 809
Explanation of Negatiye Solutions 811
Oeneral Problems involTing Complete and Incomplete Equations of the
Second Degree 811
Trinomial Equations 826
Problem of the Lights 829
UNDETERMINED COEFFICIENTS..^.....^ m- *. 885
jra i l ing \/aoea> .•■.•*.•. w.«.»*«« .*••.. •*. ••• •■•••.•.■ •«..•• ••...« •....• #•.•••.*• .•••..... u4x
Particular Cases. ^ 845
DERIVATION 847
Theorems 851
CONTENTS.
BINOMUL FORMULA ^ 856
Demonstration 857
FormaUon ofPoteerg by the RuU, 860
Demonstration of the Binomial Formula for any Exponent 864
Development of Binomials affected with Negatiye and Fractional Ex-
ponents 865
Consequences of the Binomial Formula — Square of any Polynomial... 867
Cube of any Polynomial 868
Extraction of the nth Boot of Whole Numbers and Polynomials 869
Approximate Boot of an Irrational Number to irithin a certain Vulgar
Fraction 872
Approximate Root of Whole Numbers to within a certain Decimal 878
Approximate nth Root of a Mixed Number to within a certain De-
cimal*. .• 878
Approximate nth Root of Numbers entirely decimal 874
PERMUTATIONS AND COMBINATIONS 874
Combinations 877
Permutations in which Letters are repeated 885
Partial Permutation. 886
General Examples in Permutations and Combinations 887
LOGARITHMS 890
HtmI Ptoperty. The Logarithm of the Product of any Number of
Factors taken in the same System, is equal to the Sum of the
Logarithms of those Factors 892
Second Property, The Logarithm of the Quotient arising from divi-
ding one Quantity by another, is equal to the Logaritiun of the
Dividend minus the Logarithm of the Diyisor. 898
Third Property, The Logarithm of the Power of a Number is equal
to the Exponent of the Power into the Logarithm of the Number.. 894
Fourth Property. The Logaritiim of theiJloot of a Number is equal
to the Logarithm of the Number dirided by tiie Index of the Root. 895
P^th Property. The Logarithm of the Reciproc^ of a Number is
equal to the Logarithm of the Number taken negatiyely 896
dONTBNTS.
LOQABITHMS (oontiimed).
8bah Property. The Loguitbm of any Base, taken In its own Sys-
tem, is Unily -.....- 8OT
Seoenih Fr^pviy. If we haye a Table of Logarithms oalcnlated to »
Particii^ar Base, the Logarithms of the Numbers in this Table,
^Tided by the Logarithm of a Seoond Base, taken in the first
System, will giye, as Qnotients, the Logarithms of the same Nimi-
bers in the Seoond System *»*. 897
Mferential of dx 898
Logarithmio Series 899
Common and other Logarithms found from Napierian .. ^ 401
Measure of any Modulus ^ 401
Tftble of Napierian Logarithms m...**. - • 401
Adrantages of the Common System - 404
Application of Common Logarithms 406
To find the Logarithm of a ^ven Number below 10-000 407
To find the Logarithm of a Number aboTO 10*000 406
To find a Number oorresponding to a giyen Logarithm 409
Solution of Exponential Equations by means of Logarithms....... 411
ARITHMETICAL PROGRESSION. 412
Formula for the nth Term.... .«..* 418
Formula for the First Term. 414
Formula for the Number of Terms • 416
Formula for the Sum of the Terms 416
To find the Arithmetical Mean. 418
Insertion of Means between the Extremes of a Proportion 420
GEOMETRICAL PROGRESSION 422
Formula for the nth Term 428
Formula for the Ratio of the Progression 424
Formula for the Sum of the Glories 426
An Infinite decreasing Progression 427
INEQUALITIES 481
ZU OONTEKTB.
GENERAL THEOBT OF EQUATIONS 489
Equal Yalaes. - ^ ^ 469
Deriyed Polynomials • 462
Equation of DifPerenoes m r - 467
Irrational Yalaes ••••< ^ ^ ••••••••.•« -••. 470
Newton's Method of Approximation. ..•••.•••• ,. .«, 474
Lagrange's Method of Approximation. •• ••••.- •••.••..• • 476
General Solution of Nomueiioal £qnationa*MM*M* *««•••••••••#•••••••«••• •••••• 477
DESCARTES' RULE ^ 491
ELIMINATION BETWEEN TWO EQUATIONS 07 ANT DEGREE 495
ELEMENTS OF ALGEBRA.
A&TIOUS 1. — ^A mathematical principle is a trath admitted as self-
evident, or proved by a course of reasoning called a demonstration.
Thus, it is a self-evident principle, that if equals be added to equals,
the results will be equal. But it requires a demonstration to show that
if the sum of two quantities be added to the difference of those quanti-
tiea, the result will be equal to twice tlie greater quantity.
2. Science is knowledge gained by theory or experiment, employed
in the investigation of principles. Art is the application of acquired
principles to the practical purposes of life.
3. Quantity is anything that can be measured or numbered. A
thing is said to be measured when its magnitude is expressed in terms
of the unit of measure, and is said to be numhered when this unit is
unknown or indefinite.
Thus, 12 bushels of com is a definite measure, the unit of measure
being one bushel of com. But a quantity, expressed by the number
12, or even by 12 bushels, would only be numbered, for it might be 12
bushels of gold, or 12 of flour, or 12 of anything else.
Lines are quantities, because they can be measured in terms of
yar&, feet, inches, &e. Time is quantity, being measured by hours,
minutes, seconds, &c. The unit of measure for time is generally the
second.
The operations of the mind, such as hope,. fear, joy, grief, &c., are
not quantities. For, although we speak of a great hope and a small
hope, there is no definite unit of comparison by which to measure its
magnitude.
Numbers are not quantities, but simply the agents by which to
express the abstract relations between quantities of* the same kind;
that is to say, every number may be regarded as the quotient arising
2 18 I
14 ELEMENTS OF ALQEBBA.
from dividing one quantity by another of tiie same kind^ independently
of the species to wliich they belong.
TbuB^ the number 12 may express the abstract relation between 12
inches and 1 inch, or between 12 months and 1 month; and may, there-
fore, represent the quotient arising from the diyision of the foot by the
inch, or the year by the month. And so it may express the quotient
between any two quantities whatever, provided they are of the same
kind.
4. Mathematics is the science of quantity.
Arithmetic is that branch of mathematics in which the quantities
considered are represented by numbers.
5. The word Algebra is of Arabic origin, and signifies to consolidate.
Algebra enables us to investigate arithmetical principles in a consoli-
dated, and, at the same time, general manner, and may be r^arded as
a compendious, and also universal arithmetic.
6. The quantities considered in Algebra are represented by numbers
and letters, and the operations to be performed are indicated by signs.
The numbers, letters, and signs are generally called tymbciU.
The numbers and letters that represent quantities are, for conve-
nience, most usually called quantities themselves. The student, how-
ever, should remember that they are only the representatives of
quantities.
7. The sign + is called pZtM, i. e., more^ and when prefixed to a
quantity, signifies that it is to be added to some other quantity expressed
or understood. Thus, a -\- h\& read a plus &, and signifies that & is to
be added to a. The expression + c signifies that c is to be added to
some quantity not expressed. When no sign is written before a quan-
tity, the sign -H is always understood. In the expression a •\- h^ab^
understood to be affected with the plus sign.
8. A horizontal Hne, thus, — is called mtntM, i. e. len, and when
prefixed to a quantity, signifies that it is to be substracted from some
other quantity, expressed or understood.
Thus, a — & is read a minus 6, and signifies that 5 is to be taken
from a. The expression — c signifies that c is to be taken from some
quantity not expressed.
9. A Greek cross x is called the sign of multiplication, and when
placed between quantities, indicates that they are to be multiplied
together. Thus; a X 6 is read a multiplied by 6, or simply a into h^
the sign indicating a multiplication to be performed.
ELEMENTS OP ALOSBBA. 15
10 The multiplication of qtianiities is sometimes indicated by a
point. Thus, a . h indicates that a and & are to be multiplied
together.
11. The multiplication of literal factors is usually indicated by wri-
ting them one after another, thus, ahcia the same as aX b X c. This
notation cannot be employed when numbers are used, for the product
thus expressed would be confounded with some other number. The
multiplication of 2 by 4, for instance, cannot be indicated by writing
the one after the other, because the product would be mistaJLen for 42
or 24.
12. When several terms connected by the sign +, or — ^, are to be
multiplied by a single term, the multiplication is indicated by means
of parentheses. Thus, (a + & + c) m, signifies that the sum of a, 6,
and c is to be multiplied by m. When the multiplier itself is composed
of more than one term, it is also enclosed in parenthesis. Thus, (a -f V)
(m — c) indicates that the sum of a and & is to be multiplied by the
difference of m and c. A horizontal or vertical line is also used to col-
lect terms for multiplication. Thus, ax m + n + c, indicates that the
sum of m, n, and c is to be multiplied by a. The same thing may be
indicated by a vertical bar, thus
+ma.
18. The coefficient of a quantity is a number or letter prefixed to a
quantity, showing how often it has been added to itself. Thus, instead
of writing a^a-r-a, which denotes the addition of a three times, we
abridge the notation by writing 3a. So also, lOxy, signifies the
addition of xy ten times. In like manner, mx signifies the addition
of », m times. The coefficient serves as a brief mode of indicating the
addition of a quantity to itself When no coefficient is written, 1 is
always to be understood.
14. The exponent is a small number or letter written a little above
and to the right of a number or letter, and indicates the number of
times it enters into itself as a fisictor. Thus, we write
a' instead of aa, and call the result a square.
c^ " « oaa, " " " " a cube.
tf* " " aaaa, " " " " a to the fourth power.
The exponent enables us to abbreviate the manner of indicating the
16 BLEMXNT8 OF ALGEBRA.
muhiplication of a qiumtity by itsdf. Wlien no exponent is writteu, 1
is always to be nndefBtood.
Division is denoted by three signs. The division of a by 5 may be
indicated by a -5- 6, or -j-, or a\h.
15. The sign «= is oidled the sign of equality, and is read ^^ equal
to." When placed between two quantities, it indicates that they axe
equal to each other. Thus^ a = 5^ is read a equal to &. In like man-
ner, 2 + 4 e= 6 is read 2 plus 4 equal to 6.
16. The sign ]]> is called the sign of inequality, and is read '^ greater
tlian,'' when the opening is toward the left ; and '^ less than/' when
opening is toward the right. Thus, a'^his read; a greater than h;
and c <^ m is read, c less than m.
17. A root of a quantity is a quantity which, multiplied by itself a
certain number of times, will produce the given quantity. To indicate
the extraction of a root, we use the sign VT^called the radical sign,
placing a number or letter to the left and over the sign to indicate what
root is to be extracted. Thus, ^~a denotes the square root of a;
V a" the cube root of a ; y o^the n^^ root of a.
18. The number or letter placed over the radical sign is called the
index of the radical. When no index is written, we always understand
that the square root is to be extracted. Thus, \/a means that the
square root of a is to be taken.
19. A simple quantity is one in which the letters and numbers of
which it is composed are not connected by the sign plus or minus.
Thus, a, ahf -r- and c are simple quantities or monomials. All quan-
tities not simple are compound, and called binomials when composed
of two terms; trinomials when composed of three; and polynomials
when composed of more than three. Each of the literal factors which
enter into a term is called a dimension of the term. The d^ree of a
term is the number of its dimensions. When a factor enters more than
once, its dimension is denoted by the exponent. Thus^ a' is of the
second dimension.
20. When several factors ure multiplied together, the sum of the
exponents denotes the dimension of the term. Thus, o^ is of the
second dimension, a^bc, of the fourth, al^cd, of the sixth, &c. If
the term is a fraction^ its degree is denoted by the difference between
ELEMENTS OF ALGEBRA. 17
the sums of the exponents of the numerator and denominator. Thus
is of the second degree, — j- of the first degree, &c.
c <r
21. A monomial is said to be homogeneaus, when all of its literal
factors are of the same dimension. A polynomial is said to be homo-
geneous, when all its terms arc of the same degree. Thus ab, a^b\
a'6^, are each separately homogeneous monomials; 4ca^h + 2l^c — m*, is
a homogeneous polynomial.
22. Like quantities are those which are composed of the same letters,
and which have their corresponding letters affected with the same ex-
ponents ; 3a&' -{- lOai* — 3at*, are like quantities. But 3a6' -|- XOah
— 3a'6* are unlike ; for, though the letters are like, the exponents of
these letters are different. p. 17.
23. The reciprocal of a quantity is 1 divided by that quantity. The
, .«. 1 . 1 .2 1 3 ^
reciprocal of 2 is - ; of a, - ; of - . - or -, &c.
2 a 3 2^
3
24. Quantities, affected with the plus sign, are called positive quanti-
ties, and those affected with the minus sign, are called negative quan-
tities. The student, however, should bear in mind that quantities can-
not be positive or negative in themselves, and that by these phrases, we
wish merely to signify additive and subtractive quantities.
25. To familiarize the student with the foregoing symbols, we sub-
join a few examples for practice : —
1. Express in algebraic language that twice the product of z into y,
divided by three times the product of z into to, shall be equal to 6.
2. Express that the sum of a and b, when added to their difference
is equal to twice the greater quantity a.
3. That one-third of a quantity m, multiplied by jth of a second
quantity n^ and that product increased by 100, the result will be a hun-
dred thousand.
4. Find the value of this expression, , when 6=64, c=144,
tn.n
m=10, n = l.
Find the value of the expression (x a), when wi = 4, n = 3,
m -{- n
x = 2, a = l.
7M
Find the value of the expression — „-, when m=4, n=3, rz=5,
mrn
A = 10,/ = 100.
2* B
18 ADDITION.
ADDITION.
26. Addition is the connectiDg of several terms together by the
sign plus or minus, so that they may be reduced into a single expres-
sion.
There are three distinct cases : —
CASE I.
27. When the tenns are like and have like signs.
RULE.
Add the coefficients of ike several terms together, prefix the common
sign to this sum, and lorite after it the common letter or letters, with
their primitive exponents.
Thus, -f 2a + 3a are like quantities referred to the same unit, and
may therefore be added Just as two whole numbers are added in Arith-
metic. The + 2a indicates that a is to be added tvrice to some quan-
tity not expressed, and + 3a that a is again to be added three times to
the same quantity. This addition can obviously be indicated at once
by writing + 5a instead of + 2a + 3a. Suppose, for example, that a
represented one dollar, then 2a would represent two dollars, and 3a
three dollars, and the sum, five dollars, would, of course, be represented
by 5a. In like manner, — 36 — 46 is equal to — 76, because the minus
signs indicate that the quantities represented by 36 and 46 are to be
taken from some quantity not expressed, and subtracting 76 at once is
obviously the same as first subtracting 36 and then subtracting 46. We
may then write — 36 — 46 = — 76, the minus sign before the 76 indi-
cating that it is to be taken from some quantity not expressed, and the
whole expression denoting that the subtraction of 76 is the same as the
successive subtraction of 36 and 46. This will be made plainer to the
beginner by attributing to 6 some known value, as a pound or an ounce.
EXAMPLES.
+
a
— a
+
6-
- y
-f m —
• z
— 42«
+ 2y»
+
2a
— 2a
+
26-
- 2y
+ 5m —
■ 2z
— 5««
+ ^
+
3a
3a
+
56-
- 4y
-f 8m—
Sz
— 9«*
+ 7y»
+
4a
10a
— 4a
-h 96-
-1-176-
- 8y
-15y
-hlOm —
+ 24m —
• 7z
13«
— 122«
+ 8y»
+ ■
10a
— 30s«
H-22y»
ADDITION. 10
CASEn.
28. When tlie quantities are like, but affected witli unlike signs.
RULE.
Add the guantitiez affected with the positive tign hy the last rule, then
add those affected with the negative sign in like manner. Subtract the
smaller of the coefficients of those sums from the greater, annex the
common letter or letters to the difference, and prefix the sign of the
greater sum.
If we were required to add + 5a — 6a -f 4a — a together, we could^
for the reasons already given, write + 9a for the positive terms, and 7a
for the negative terms. By this, we must understand that 9a has to be
added to some quantity not expressed, and that 7a has to be subtracted
from the sum of 9a^ and this unexpressed quantity. Now to take 7a
from 9a, and add the remainder to the unexpressed quantity is plainly
the same as taking 7a from the sum of 9a, and the unexpressed quan-
tity. But the difference between 9a and 7a is 2a, hence the sum of
4- 5a — 6a -f 4a — a is + 2a. The plus 2a denotes that 2a has to
be added to an unexpressed quantity.
If we were required to add — 5a + 6a — 4a -f ^^ together, we could,
as has been shown, collect the negative terms into a single term, — 9a,
and the positive terms into a single term +7a. We would then be
required to take 9a from the sum of 7a and some unexpressed quantity,
or, which would be the same thing, to take 9a from 7a, and add the
remainder to the unexpressed quantity. But 9a is made up of 7a
added to 2a, and when, therefore, we subtract 9a fh)m 7a, the 7a'8
will strike each other out, and there will still remain 2a to be sub-
tracted. The required and unexecuted subtraction is indicated by the
minus sign before the 2a, and the true sum of — 5a + 6a — 4a 4- a
is therefore — 2a.
EXAMPLES.
1. Add 2 5 — 66 + a6 — 26 together.
Ans, +ab — 66 or (-|- a — 6) 6.
2. Add 4c -I- 3c — mc + ne together.
Ans, (J + n — m) c.
8. Add xy + 2xy -f- hxy — pxy together.
Ans. (3-1-6 — J)) 5cy.
20 ADDITION.
4. Add 4a — 7a — 3a — 10a — 12a tc^ther.
Ans. — 28a.
6. Add 4aa; — 7ax — 3a« — lOoa: — 12aa5 together.
Ans. — 28a2;.
CASE m.
29. When the quantities are similar and dissimilar; and have like
and unlike signs.
RULE.
Add all the sets of similar terms hi/ the last two rtdeSy and write
their sums one after another^ and connect with them all the single terms
with' their own signs.
Since the letters are always the representatives of quantities, it is
obvious that the quantities represented by dissimilar terms cannot be
added into one sum. Thus, 4a + Ah neither make 8a nor %h. The
quantity a might represent a year, and the quantity h a pound ; then
4a would represent 4 years, and 46 would represent 4 pounds, and the
addition ought neither to give 8 years nor 8 pounds. Hence, we can
only reduce the similar terms in sets, and connect the results with their
appropriate signs.
EXAMPLES.
3m«
— n" ay
4- 7x bax
+ y
2nx-\- w
3x«
+ wi" «
+y a^
+ 4y
2w -f bnx
2m»
+ a* ocy
4-3a5 As^
+ %
6nx-\-s
6m' + 4x*— n» '2ary+10x+a+y 5x*+5aar+14y 13nx-{-Sw-\-8
30. It often facilitates an algebraic operation to arrange the terms in
a certain order. In addition, this arrangement is effected by placing
all the like terms beneath one another.
7a* 4 6ay + 5« + 4w 7a* + 4w + 6ay + bz
w +4z + 8a:y+ 12a' may be written 12a' -f w + Sxi/ + 4z
Zz + 9xy+ bw + 100a' 100a'+ 5w 4- O.T.y -f> 3g
119a'+10w + 23yy + 12«
31. Sometimes the addition of a compound quantity to a single quan-
tity, or to another compound quantity, is not actually performed, but
indicated by the parenthesis (), or vinculum. Thus, 46 -|- (6 — c), or
45 ^ I — Cj indicate that h — c is to be added to 46.
BUBTBAOTION. 21
JRemarks,
32. It will be noticed that the tenn addition in Algebra^ is used in
an extended sense, the operation being oflen arithmetical substraction.
The addition of a negative quantity is, in fact, the same as the subtrac-
tion of the same quantity regarded as positive. The use of negative
quantities in Algebra constitutes one of its marked differences from
Arithmetic, in which the numbers employed are always supposed to be
positive. By reference to the examples, another remarkable distinction
between Algebra and Arithmetic will be observed ; each sum indicates
what quantities were added together, whilst an arithmetical sum con-
tains no trace of the numbers employed. Thus, the sum of 10 and 5
is 15, but the result does not point out the numbers that were added,
for 15 might proceed from the addition of 14 and 1, 12 and 3, 13
and 2, &c.
It will be seen hereafter, that in all algebraical operations, the
result contains some, if not all, the quantities employed in the inves-
tigations.
SUBTRACTION.
33. Subtraction is taking the difference between two or more
quantities, and may be regarded as the undoing of a previous addition.
If we were required to subtract + 6 — 4, or 2 from 12, the result
plainly ought to be 10. But if we write 6 — 4 beneath the 12, and
perform the subtraction of each term separately, the subtraction of 6
from 12 will give a remainder 6, which is too small by 4, because we
were not required to take 6, but 2 from 12. We can only correct the
error by adding + 4, and, hence, we have 12 — ( + 6 — 4) = 12 —
6 -h 4 sa 10, as before. We observe, in the last result, that the signs
of 6 and 4 have both been changed.
Again, let it be required to subtract + h — c from a. The sub-
traction of b from a will be indicated by writing a — b, but this re-
mainder is too small by c, because we were not required to subtract h
itself from a, but what remained of b after it was reduced by c. The
error can only be corrected by adding plus c to the result. Henoe>
a — (-h 6 — c)=s a — 6-fc.
22 BUBTBACTION.
In thifl general example, we see that all the signs of the subtrahend
have been changed, and theui it has been added, as in addition.
Hence^ we have the following
RULE.
Conceive the ngn$ ofaUthe terms of ike subtrahend to be changed^
and then add up these terms cu in addition.
EXAMPLES.
— 6a + 45 Sy/x + 5a» 7a? t m 2v a + ax
+ 4a— 6 6>/x— 8a* 7a? -f 2m 4v^+ Gax
— 10a 4- 56 —sTx-^-lSa" 0— m —2y/~a — ^
From v/F+ 6 — 10 +8a*+76y +18a+19:c +lln
Take cf— 20+ 9x + 5n— 2^/x^- 364- 4a» + » + 2«
3>/ar— 26 + lOx +4a« +^180 +Qn—d—z—28 +76y+10
In the third example we have written zero as the difference between
7a? and 7a?, but it is more usual to denote zero by a blank.
From 3xy + 14 y/1/ -{• z + 2n — w
Take 3a-y + « — 11 -f 11 >/y^ 27i + to
11 + 3^^+ ^n — 2w
From 12a" + 46a; — ca? + «i* + 46" —<?a? + ns
Take 46« — c«x" — 66a? + ca? — m* —Sns + 106" —
106x— 2cx" 4- 2fn* + 4n« — 106" + try
34. The remainder added to the subtrahend ought to be equal to the
minuend, and the result, therefore, can be verified, as in Arithmetic.
It will be seen that an Algebraic subtraction does not necessarily im-
ply diminution, and that the remainder may not only exceed the sub-
trahend and minuend separately, but may be equal to their arithmetical
sum. In general, the subtraction of a negative quantity is equivalent
to the addition of the same quantity taken positively. Thus, — 6 taken
from a gives a + 6 for a remainder. This subject may be illustrated
by a simple example : Suppose a to represent the value of an estate
exclusive of its liabilities, and 6 a mortgage upon that estate, then a — b
My 4- 12a"
SUBTRACTION. 28
will represent the actual value of the estate. Now, suppose some friend
of the owner should determine to cancel his deht ; he could do this,
either by removing the mortgage (in algebraical language, taking away
— l>)j or by giving him a sum of money equal to the debt, that is,
a sum represented by + h. So we see that, taking away — 5 is equi-
valent to adding -f h.
35. Quantities are considered negative, when opposed in character or
direction to other quantities of the same kind, that are assumed to be
positive.
Thus, if a ship leave port with the intention of sailing due north, but
encounters adverse winds, and is driven south part of the time; to get
the distance passed over north, we must obviously subtract the distance
sailed over south. If then the direction north be considered positive,
the direction south must be considered negative. Suppose the vessel
sails first day 100 miles north to 10 south, second day 80 miles north
and 30 south, the entire distance passed over in a northern direction
wUl be expressed by 100 + 80 — 10 — 30 = 140.
If a man agree to labor for a dollar per day, and to forfeit half a dollar
for every idle day, and he labor 4 days and is idle 2, his wages will be
4 X 1 — 2 X }, or 4 X 1 + 2 ( — i) = 3. We see that we have re-
garded as negative, cither the forfeiture as opposed to gain, or the idle
days as opposed to the working days.
If a man's age be now thirty years, his age four years hence will be
expressed by 30 + 4 ; his age four years ago by 30 — 4. Here future
time being positive, past time is negative. If a steamboat sail with a
velocity of 10 miles per hour, and encounter a head wind that would
carry it back at the rate of 8 miles per hour, then its rate of advance
will be expressed by 10 — 8, or 2 miles per hour. But if it be carried
back at the rate of 12 miles per hour, then its rate of advance will be
expressed by 10 — 12, or — 2 miles per hour. It will then plainly be
carried back, and the minus indicates a change of direction.
Distance, when estimated as positive in one direction, is considered nega-
tive in the opposite direction. Thus, let the distance AB = n and BC = m,
C
i J. — , the distance BC being considered positive on the
A B '
right of B. Then AC == m + w. Now suppose the distance BC be esti-
mated on the left of B, the point C falling between B and A, then
AC = m — n. We see that the distance BC, which was regarded aa
positive on the right of B, became negative when estimated on the left.
24 MUIiTIPLI CATION.
In general, the minus sign may be regarded as always indicating
either that a quantity has changed its character or direction, or that it
is just the opposite of something else of the same kind that is con*
sidered positive.
MULTIPLICATION.
86. Multiplication is a short method of adding the multiplicand
to itself as many times as there are units in the multiplier. Thus, to
multiply a by 6 is to add a to itself b times, and since the addition of
a to itself twice is 2a, three times 3a, &c., the addition of a to itself b
times will be ba. Hence the product of a by 6 will be ba or ab, for it
plainly matters not in what order we writes the factors.
There are three cases : —
CASE L
87. When both multiplicand and multiplier are simple quantities.
Before giving a rule for the multiplication, it will be necessary to
show that the product of simple quantities having like signs, both plus
or both minus, is always positive, and that the product is always nega-
tive when they have unlike signs. The product of -f a by + 5 is
plainly + a&, because plus a added to itself b times must, of course, be
positive.
And — a by + 6 must be negative, because — a added to itself any
number of times must, of course, retain its sign. But — a by — b
will give + ab, for the b having a minus sign before it, indicates the
reverse of what it did before, and therefore denotes that — a is to be
subtracted from itself b times. But we have seen that the subtraction
of — a is the same as the addition of + a; in like manner, the subtrac-
tion of — a twice is the same as the addition of + 2a ; the subtraction
of — a three times the same as the addition of + 8a ; and the subtraction
of — a, b times, the same as the addition of + a, + b times, or -f ab.
Hence, the product — a by — 6 is -f- ab, and we see that like signs
produce plus, and unlike signs, minus.
Let it now be required to multiply a* by a*. The exponents indicate
that a enters twice as a factor in a', and four times as a factor in a,
hence; it will enter 6 times as a factor in the result.
MUIiTIPI^ICATION. 26
Hence, a' x a^ = aaaaaa = a', and we see tbat the mnltiplicaiion is
effected bj adding the exponents of the same letter.
To multiply a' by ah is the same as multiplying a' by a, and then
multiplying the result by h. But to multiply a* by a, we have only to
add the exponents of a' and a, hence the result is o^. Now multiply
by hy and we have a*6.
The multiplication is effected by adding the exponents of like letters,
and writing after the result the letter which is not common to the mul-
tiplicand and multiplier, with its primitive exponent. In like manner
o* X oh^ will be a*}?. To multiply a'r by ahy is the same as multiplying
(j^c by a, and then the result by h. But a^c by a, as we have seen, gives
a'c, and that product by h, will plainly be a?cb. And the multiplica-
tion is again effected by adding the exponents of like letters, and writing
after the result the letters, which are not common, affected with their
primitive exponents. The same reasoning can be extended to any num-
ber of factors, the multiplication being effected in every instance^ by
adding the exponents of like letters, and writing after the result, the
letters not common, with their primitive exponents. Thus, a^cd by ah
gives c^cdhj and c^cd by ahm gives a^cdhm. We have taken mono-
mials whose coefficients were unity. If we were required to multiply
a by 2h, the product of a by 5 has to be multiplied by 2, but the pro-
duct a by 2) is ah, hence the result is 2ah. If we were required to
multiply 2a by &, then 2a has to be added to itself ^ many times as
there are units in h, but to add 2a to itself twice gives 2a x 2 or 4a,
to add it three times gives 2a X 3, or 6a, and to add it h times gives
2a X h, or 2ah. In like manner, if we were required to multiply 2a
by 45, then 2a has to be added to itself 45 times, and the result will
plainly be 2a x 45, or Sah : the same result that we would get by
multiplpng the literal fitctors together, and prefixing the product of
the eoefficient for a new coefficient. Hence we have for the multipli-
cation of monomials, the following
RULE.
38. Multiply the coefficients together for a new coefficient, and write
after it all the literal factors, common to the ttoo monomials, affected
with exponents equal to the sum of the exponents in the multiplicand
and multiplier, and those, which are not common, with their primitive
exponents. If the monomial* have like signs, give the plus sign to the
prod^ict; if they have unlike signs, give the minus sign to the product.
3
26 MULTIPLIOATION.
EXAMPLES
1. a'6c
by aZ>c.
-4n«. a'6V.
2. —a*be
by afec.
Ans, — a'fcV
3. dl'hc
by — ahc.
-4?i«. — a'6V.
4. 2a^hc
by aftc.
i4 w«. 2a»i»«c».
6. 2a«6c
by ahc.
^n«. — 2a»6V.
6. 2a«6c
by — flic.
jIw*. — 2a^W.
7. a«6c
by 2ahc,
^n». 2a»6V.
8. a'6c
by — 2ahc,
^rwr. 2a>Z^»c«.
9. a"&»
bya»+'6"+».
^ws. a'-'-'^*"^*.
10. a+"6-
by a"-»Z>"-'
jln«. d^'h^\
11. a-"6-»c-»
by a-'ft-V.
Ans. a-^-'-^cr" '.
12. 4a-"'6"c
by— 8a"'6-»c-\
^n». — 32a»6*-*6'-».
13. 4a'6V
by — 8a^^>V
Jns. 32a'6V.
14. 6a-»6-»c-»
by 7a-'*6-V-\
^ns. 42a-*'6-'c-^.
15. 3a«^V
by 9a-*6-*c-».
^7W. 21a%*c\
16. oTf^
by 4a?y25.
Ans. 4a;'" '^■+'«H->^,.
17. a:"^«c
by ac.
-4w8. + T^yzcu?.
18. iryc
by xy.
Ans. x^ yc.
19. a^yc
by x/.
Ans. a^^^c^
20. ajP-y
by icV .
Ans. xP+yc.
21. T^-yc''
by c'x'.
-4ns. x^ ^ yc.
39. We see, from the last four examples, that we can make any
change in tbe position of the exponents, provided we retain the factors,
and get the same result.
We see, from examples 13, 14, and 15, that when the factors of the
multiplicand and multiplier are homogeneous, the factors of the result
will also be homogeneous.
Example 11 shows that this will not be true when the multiplicand is
alone made up of homogeneous factors ; and Example 1 shows that it
will not be true when the multiplier alone is homogeneous.
40. Any number of monomials may be multiplied together, in accord-
ance with the preceding principles.
MULTIPLICATION 27
EXAMPLES.
1. a'c X ahx hd X mn. Ans. c^l^dcmn.
2. — a'c X — abx bd X mn. Ans, a^I^dcmn,
3. — a*c X — ahx — hd X mn. Ans. — c^h^dcmn.
4. — a*cX — ahx hd x — mn. Ans. — a?l^dcmn.
We see^ that when the monomials are all positive, the result will be
positive ; and when the negative monomials are even in number, the
result will also be positive. But when an odd number of negative mo-
nomials are multiplied together, as in the fourth Example, or when an
odd number of negative monomials are multiplied by a positive mono-
mial, as in Example 3, the result will be negative. The result would
also be negative, if an odd number of negative monomial &ctors was
multiplied into any number of positive monomial factors.
CASE n.
41. When the multiplicand is a compound quantity and the multi-
plier a monomial.
The multiplicand is to be repeated as many times as there are
units in the multiplier ; each term of the multiplicand is then to be
multiplied by the multiplier, and the partial products to be connected
with their appropriate signs. We might assume, what has already
been proved for monomials, that the product of a positive quantity by a
positive, or of a negative quantity by a negative, gives a positive result;
and that the product of a positive quantity by a negative gives a nega-
tive result. But we can demonstrate this rule more rigorously : Let it
be required to multiply a — a by -f c. We know that a — a is zero,
and that zero repeated c times must still be zero. Hence, the product
of a — a by -f c must be zero. But -h a by + c gives -f ac for a
product, and, therefore, the product of — a by -f c must be — ac, in
order to cancel the first product. Hence, the product of a negative
quantity by a positive gives a negative result. Or, to express the
whole algebraically,
a — a = o
-f c
-f ac — ac=^
28
MULTIPLICATION.
Let it now be required to multiply a — a by — c. The multipli-
cand being zero, the product must be zero. But, from what has just
been shown, the product of + a and — c is — ac, hence, the product
of — o by — c must be + ac, to destroy the first product. Or, in
other words, the product of a negative quantity by a native mu3t be
effected with the positive sign.
Algebraically, a — a = o
— c
— ac •{• ac
We conclude that the product of a positive quantity by a positive, and
of a negative quantity by a negative, is positive ; and that the product
of a positive quantity by a negative is negative. Briefly, we say like
signs give a positive result, and unlike, a negative result
RULE.
Multiply each, term of the multiplicand by the multiplier, and otm-
nect the results with their appropriate signs.
EXAMPLES.
a' + 6 by a.
a* — h by a.
— a' + 5 by a.
— a' — b by — a.
— a' + ft by — a.
a^b + c + d by a.
a'b — c + d by a.
a'fe + c — d by a.
— a^b + r. + d by a.
— a^b — c + rf by o.
— a'b — e — d by a.
— a*b — c^^d by — a.
— a*b — c + d by — a.
— a^b + c + d by — a.
a*b + c + d by — a.
m^n + a + 2c by «.
m*w + a 4- 2c by 3c
m'» + a + 2c by 4mn.
8m — 6a + llcby2c».
Ans. a' + ah.
Ans. a' — ab.
Ans. — a' + ab.
Ans. a* + ab.
Ans. a' — ab.
Ans. a^b + ac -f- ad.
Ans. a^b — ac -\- ad.
Ans. a^b -{- ac — ad.
Ans. — a*b + ac + ad.
Ans. — a'fe — ac '^- ad.
Ans. — a'6 — ac — ad.
Ans. a*b + ac -f- a<f.
Ans. a^b -^ac^^ ad.
Ans. a*b — ac — ad.
Ans. — d^b^-^ac — ad.
Ans. w^ns -f <w + 2cs.
Ans. Scm^n + Sac + 6c*
Ans. 4m^n^ + 4amn + Bmnc.
Ans. 16mc»— 12ac» + 22c^
MULTIPLICATION. 29
CASEIIL
42. When the multiplicand and multiplier are both compound quan-
tities.
The multiplicand, as in the two preceding cases, is to be repeated as
many times as there are units in the multiplier. The multiplicand and
multiplier will, in general, be made up of some positive and some nega-
tive terms. Let a denote the sum of all the positive terms in the mul-
tiplicand, and b the sum of all the negative terms. Let c denote the
sum of the positive terms in the multiplier, and d the sum of all the
negative terms. We write the multiplier beneath the multiplicand
and multiply all the terms of the one by all the terms of the other.
Thus,
a
c
h
— d
etc
— be
— ad + bd
ac —
be — ad -^ bd
To explain this result, let us drop for a moment the consideration of
— d, then a — b must be repeated c times. From what has been
shown, the result of this multiplication will be ac — be. But we were
not required to multiply a — bhy c alone, but by c after it had been
diminished by d; hence, the result, ac — ^, is too great by a — b
taken d times. To correct the error, then, we must subtract the pro-
duct of a — bhy d from the first product ac — be. The product of
a — bhj + d, from what has been shown, will be -H ad — bdy and
to indicate that this must be subtracted, we write it in parenthesis with
the minus sign before it. The whole result will be ac — be — {+ ad
— bd") = ac — be — ad + bd, since the signs of the quantities sub-
tracted must be changed. By examining the result, we will observe, as
before, that the product of quantities affected with like signs is positive,
and that of quantities affected with unlike signs is minus.
43. It is found most convenient to arrange both polynomials with
reference to the highest or lowest exponent of the same letter. Thus,
7? + a^ -^ X + a\& arranged with reference to the highest exponent
of x; and a-{-x-\'Q?-\-7?i& arranged with reference to the lowest
exponent of the same letter. In these expressions, x is supposed to
enter to the zero power in the term a, as will be explained under the
head of IHvision.
3*
30
MVLTIPLIOATION.
BULE.
44. Arrange the multiplicand and midttplter with reference to the
hiyhezt or lowest exponent of the tame letter (if they have a common
letter')^ and then multiply each term of the one polynomial by each term
of the other polynomial, beginning on the left. Set down the result of
the multiplication of the second term of the multiplier under that of the
first term, only removed one place further to the right, and the result
of the multiplication of the third term of the multiplier under thai of
the second, and continue the operation until the multiplication is com-
plete. Then reduce the whole result to the simplest form.
EXAMPLES.
1. xi-h a
X — a
— ax-
a'
a"
2. x^ + xy -\- a
X -\-y
x^ + a^y + ax
H- a^y + xy*+ ay
a^ + 2x*y 4- ox -f xy" + ay.
x
X
ft. X* + 2xy + ^
X — y
x* + 2x2^ + yx
— a^y — 2^x
y
x' + x*^— /x — y*
X* + o — a" = X* —
3. a^ -|- ax* — X
x" + ax + 1
x5 + ax* — X*
+ ax* + ax* — ax*
+ x»+ ax«-
x5 + 2ax* + ox* + o —
45. It will be observed in the above examples, that by arrangiDg
with reference to a certain letter^ and by removing each result one
place further to the right than the preceding, the terms which reduce
fall immediately the one below the other. And it is to save the trouble
of looking for the terms which reduce, that this arrangement and sys-
tem are made. It will also be noticed that in eveiy product there are
two terms which do not reduce with other terms, viz., those which
result from the multiplication of the quantities affected with the highest
and lowest exponents of the arranged letter.
In the first example, a^ and — a' are the irreducible terms, in the
second, x^ and ay.
MULTIPLICATION.
81
EXAMPLES.
6. Multiply x + y by aj + y.
Ans. a? + 2ay + y*.
7. Multiply X + y by 05 — y.
-4n». «" — y«.
8. Multiply a^ + 2xy + y* by + a; + y.
Ans. x* + Sx'y + 3y*a; + y*.
9. Multiply 7? — 2xy + y" by x + y.
Ans. 7? — 7?y — xy* + y".
10. Multiply X* — y* by cc« + y».
-4n5. X® + y*^* — x'y* — y*.
11. Multiply a' + 2ax + x" by x + a.
Ans. x* + Sax* + 3a'x+ a*.
12. Multiply x* + tt* + 3«^ + ^^a:" by x + o.
Ans. X* + 4ax* + 6a*x' + ^o^-^ + «*•
13. Multiply 7x* + 2a by 3y*-
Ans. 21x*y* + 6ay*.
■*-/
14. Multiply 4x* + y* by 4x* — y*.
uin.^. 16x^
16. Multiply x* + ax' + a'x + a' by 2a + 2x.
Ans. 2x* + 4ax' + 4a»x« + 4a'x + 2a^
16. Multiply x* + xy + y» by 7x + 7y.
Ans. 7r» + 14x*y + 14xy« + 7y».
17. Multiply X* + ox* + aV + a^x + a* by 4x + 4a.
Ans. 4x* + 8ax* + 8aV + 8o'x» + 8a*x + 4a'.
18. Multiply 2x* + 2ax + 2a' by 3x + 3y.
Ans. Gx* + 6ax* + Cx'y + 6a^x + 6axy + ^a^y.
19. Multiply 2a" — 26« by 2a + 25.
Ans. 4a» + 4a«6 — 4a6» — W.
^0. Multiply 2x + 3a + 46 by y + z-
Ans. 2xy + 3ay + 4fty -f 2ocz + Saz + 4hz.
21. Multiply 2x + 3a + 46 by 2x — 2y.
Ans. 4x' — 4xy + 6ax + 86x — 6ay — 86y.
22. Multiply 4y» + 4x* + imn by 2x + 2y + 2z.
Ans. 8y» + 8yx* 4- 8ymn + 8.c* + 8.ry« + 8x7nn + iy'z + 8x'«
-{' Sfnnz.
MULTIPLICATION.
Remarks,
46. It will be seen, by iDspecting the above results, tbat when the
two polynomials are both homogeneous, the product is also homoge-
neous, and of a degree equal to the sum of the dimensions of the mul-
tiplicand and multiplier. Thus, in Example 7th, the first polynomial
is homogeneous, and of the first degree ; and the second is also homo-
geneous of the first degree, the product is homogeneous of the second
degree. In Example 8th, the multiplier is homogeneous of the second
degree, and the multiplier is homogeneous of the first degree; the product
is homogeneous of the third degree. The same may be noticed in
several other examples.
2d. We notice also that when the coefficients are the same in each
term of the multiplicand, and the same in each term of the multiplier,
and all the terms of both polynomials are positive, that the sum of the
coefficients in the product will be equal to the product arising from mul-
tiplying the sum of the coefficients in one polynomial by the sum of the
coefficients in the other. Thus, in Example 16, each coefficient of the
multiplicand is 1; and each coefficient of the multiplier 7. The sum of
the coefficients of the multiplicand is 3, and that of the multiplier 14 :
the sum of the coefficients in the product is 8 X 14 = 42. The same
law may be noticed in Examples 15, 18, and 22.
47. Examples 10 and 19 show that when the multiplicand is com-
posed of the difference of two terms, whose coefficients are equal, the
algebraic sum of the coefficients in the product is zero. Examples 7
and 21 show that this sum is also zero when the multiplier is composed
of two terms with contrary signs and equal coefficients.
3d. It has already been remarked that there are always two terms,
which do not reduce with any other terms. We can only reduce similar
terms, and when the two polynomials have been arranged with respect
to a certain letter, the products of the extreme terms are dissimilar to
the other partial products. The whole process of division of polynomials
is based upon this fact, and it ought to be remembered. By attending
to the above laws in regard to the product, we can often by a simple
inspection detect errors in the multiplication.
There are three theorems of great importance, which must be com-
mitted to memory.
MULTIPLICATION. 33:
Theorem I.
48. The square of the sum of two quantities is equal to the square
of the first, plus the double product of the first by the second, plus the
square of the second.
Let X denote the first quantity, and a the second, then x-\- a^ss^ sum
and (x -f afssi{x 4- a) (a; -f a), which, by performing the multiplica-
tion, will be found equal to a:* + 2ax -f a*.
Hence {x + a)*= x' -f 2aa; H- a', as enunciated.
So (10+ 5y = 10*+20x5+6* = 100+ 100 + 25 = 225.
And (40 + 6)« = 40« + 80 X 6 + 6« = 1600 + 480 + 36 = 2116.
In like manner (x" + a")* = x*" + 2x"a'" + a*".
The rule may be extended to binomials of any form.
Thus, (3a + hY = (3a)* + 6a6 + 6' = 9a* + 6a6 + 6^
Qx + 5yy = (7x)«+ 14x (5^) + {hyf = 49x» + 70xy + 25/.
A polynomial may be squared by the same formida.
Let it be required to square x + a + y.
Make x + a = «.
Then x+a + ya=2; + y and (x + a + y)' = (2 + y)* = «* + 2zy
Now substitute for z its value x + a, and we have (x + a + y)* =
(x + a)» + 2 (x + a) y + y = x« + 2ax + X* + 2xy + 2ay + 3^.
Kequired the square of 2y + x + m + «.
Let 2y + X = «, and m + w = «.
Then (2y + x + m + n)* =(«+«)« = 2« + 2» +«' = (2j^ + x)»
+ 2 (2y + x) (m + n) + (w + n)' =s 4^* + 4yx + x* + 4ym ^- 4y?*
+ 2xm + 2xn + m* + 2»mn + n*.
Required the square of 8y + 4x + m + 2n+ 5.
Represent the first three terms by a single letter, and the lajst two
also by a single letter, and proceed as before.
Theorem II.
49. The square of the difference of two quantities is equal to the
square of the first, minus the double product of the first by the second,
plus the square of the second.
Let X and a denote the quantities, then x — as difference.
And (x — of = (x — a) (x — a) == x' — 2ax + a", as enunciated.
So(10 — 5/«10='— 20x 5+ (5)* = 100 — 100 + 25 = 25.
And (40 — 6)« « (40)* — 80 x 6 + 6« = 1600—480 + 86 = 1156.
c
M MULTIPLICATION.
In like manner (x" — a")" = a:*" — 2x'"a'" + a"".
Bequired the square of 2x — a. Ans. 4x* — 4ax + a*.
Kequired the square of 3a; + b — a.
Ans.(Bx+ hy — 2(Sx + b)a+ a» = 9x»+65x+ h'—6ax — 2ab
+ a".
Required square of a::* — a*. -4n«. x* — 2a:*a' + a*.
Theorem III.
50. The sum of two quantities multiplied by the diiFercncc of the
same quantities^ is equal to the difference of their squares.
Let X Kud a denote their quantities.
Then x 4- a = sum, and x — a =s difference, and (x + a) (x — a)
= X* — a', by performing the multiplication indicated.
So (10 + 6) (10 — 5) = (10)« — (5)» = 100 — 25 = 75.
And (40 + 6) (40 — 6) = (40)* — (6)« = 1600 — 36 = 1564.
Multiply (4a + 6) by (4a — 6). Ajis. 16a» — 6^
Multiply 7a -h b + c by 7a + b — c.
Ans. (7a + 6)« — c^ = 49a« + Uab + i« — c«.
Multiply x + yH-« + m by xH-y-f« — m.
Ans. (x+j^ + «)' — m» = x« + ^+««+2x^ + 2x« + 2yz — m«.
Multiply x + y+^+wibyx-j-y — z — m.
Ans, (x + t/)' — (2 -f- my =s x* + 2x^ + y" — s^ — 2mz — m*
Bemarks.
51. If we omit the exponents of the extreme terms in the expres-
sion, x' + 2ax + a*, and connect these terms with the exponents so
omitted, by the sign of the middle term 2a j^, we have x + a, the bino-
mial, which squared gave x* + 2ax + a'. In like manner, if we omit
the exponents of x" and a' in the expression, x" — 2ox -f a*, and con-
nect them with their exponents omitted by the sign of the middle term,
we have x — a, the binomial, which squared gave x* — 2ax -f x*.
In like manner, if wo have given the difference of two squares, we
can readily determine the quantities which, by being multiplied together,
gave this difference.
Thus m* — n» = (m + n) (w — 71), and / ' — «' = (/• -\. s) (r — «).
Omit the exponents, connect the terms by the sign plus, and we have
the sum of the two quantities ; omit and connect by the sign minus, and
FAGTORINQ POLYNOMIALS. j85
we have the difference. And the prodnct of the sum by the difference,
is the difference of the squares. These remarks are preliminary to an
important subject.
FACTORING POLYNOMIALS.
52. It is often a matter of great importance to resolve a polynomial
into its factors. The reduction of expressions can often be effected in
this way, and in no other. Practice alone can make the student expert
in the resolution of an algebraic expression into its factors. A few rules
may, however, assist the beginner.
53. 1st. Look out for the terms which have a common factor, and
write them in parenthesis, as a multiplier of that factor ; nei;t look for
another set of terms also having a common factor, and write them in
like manner; proceed thus until all the terms are taken up. Observe
whether the parenthetical expressions are the same, if so, multiply the
common parenthesb into the algebraic sum of the terms to which it
serves as a coefficient.
EXAMPLES.
1. 6a5 + fta + ca; + ca = 5 (x + a) + c (a; + a) =s (x + a) (^ + 0*
2. hx + ha — ex — ca ^s^h (x + a) -^r (x -{■ a) ( — c) =s (x + a) (h
-e).
3. — hx — ha +cx+ca=i ( — h)(x+ a) + (x+ a)c=s{x + a) (c
-6).
4. bx + ex -{- ha+ dx -\' da -{- ca sss (x + a)h -\- (x-\-a)d-\' (x + a)
d^(x+a) (6+c+(f).
5. bx+ ex + ba — dx — c7a + ca = (aj + a) (6 + c — d),
6. x" + ax -{- bx + ah s= (x + a") X -\-(x + a')b = (x + a") (x + b).
7. ac* — ox + ftx — a6 = (x — a) (x + 6).
8. a^ — ax -{• bx — ab -{■ a^ — ax' = (x — a) (x + b + 7?),
9. 7? — ax + 6x — a& _f- mx* — awx" = (x — a) (x + 5 -f ma^.
10. 7? — ax -f bx — ab + amot^ — ma? ^ (x — a) (x + 5 — mx").
54. 2d. After having found a common factor or parenthesis, see
whether that factor may not admit of farther reduction. (Articles 48,
49, and 50.)
3G FACTORING POLYNOMIALS.
EXAMPLES.
1. fcn« + 26n + 6 + cn« + 2c» + c = 2> (n* + 2« + 1) + c (n* + 2n
+ 1) = (6 + (n« + 2n + 1) =: (6 + c) (n + 1)«. (Article 48.)
2. 67i« — 2&71 + 6 + «i« — 2c» + c 5= (i» + c) (n — 1)*. (Art. 49).
3. am* + bm*—an* — 6?i» = (a + Z*) (m«— 7i«) = (a + i») (m +
fi) (m — n). (Article 50).
4. am* — 6m' — an' + 5n' = (a — 6) (m + n) (m — n).
5. n' + 2n' + n = 71 (n + 1)'.
6. n* + 2n' == n' (?i + 2). Admits of no lower reduction.
7. n»— 2n' + n = n (n - 1)'.
8. n' — 2n* + n + mn* — 2mn + m ^^ (n -^ m) (n — 1)'.
9. n' — 2n'H-n — mn* + 2mn — m = (n— m) (n— 1)'.
10. — n' + 2n' — n + mn« — 2m» -f «*=■ (m — n) (it — 1)«.
11. — n* -f 2n'*— n — mn' + 2m» — m = — (n + m) (» — !)*•
12. TO« — n'— 2cn — c' = m« — (n + c)'.
We have now the differences of two squares^ and to apply the formula
a^ — a' =s (aj + a) (x — a). (Art. 50).
We see that m' =i a;', or m =s x, and (n -f- c)* = a', or « + c ^ a.
Hence (x + a) = (wi + n + c), and (x — a) as (m — n — c).
Therefore m' — (n -{• c)' =z (m + n + c) ("* — ^ — 0«
13. m' — n' + 2rn — c' sas m' — (n — c)* = (m + n — c) (m + c
— n). In this Example m =s a?, and n — c =b a.
14. m' + 26m + 6'— w' — 2c7i — n« = (m + 6)' — (n + c)« = (wi
4- 6 4- n + c) X (m -f 6 — n — c).
55. 3d. Expressions may sometimes be thrown into factors by an
artifice, when they do not at first glance appear to admit of resolution
into factors. One of the simplest contrivances to effect a decomposition
into factors is the addition to, and subtraction of the same quantity
from the given expression, which operation does not, of course, alter its
value.
EXAMPLES.
1. a' — 2x — 15 = 2;' — 2x + l — 15— l«a:* — 2a:-f 1— 16
= (x — 1)' — 16 = (x — 1)' — (4)' = (x — 1 + 4) (x— 1 — 4).
Art. 50. = (x + 3) (x — 5).
FAGTOBINQ POLYNOMIALS. 37
In this example; the decomposition was effected hy adding + 1, and
subtracting the same from the given expression. The first three terms
of the equivalent expression were thus made a perfect square = (x — 1'),
and the whole was made the difference between two squares.
2. a:* + 4x — 12 « x* + 4x + 4 — 12 — 4 » (x + 2)« — (4)* «
(x + 6) (x — 2).
3. x« + 2x — 8 = x« + 2x + 1 — 8 — 1 = (x + 1)* — (3)« «
(x + 4)(x— 2).
4. a,-" + 4x — 21 = x« + 4x + 4 — 21 — 4 = (x + 2)«— (5/ »
(x + 7) (x — 3).
5. x« + 8x — 48 = x» 4- 8x + 16 — 48 — 16 =« (x + 4)« — (8/
= (x + 12) (x — 4).
6. x» -f-lOx + 24 = x«+ lOx + 25 + 24 — 25 = (x + 5)» — (1)«
= (x-f6)(x + 4).
7. X* + 8x + 12 = x» + 8x +16 + 12 — 16 = (x + 4)» — (2)«
= (x + 6) (X + 2).
8. X* + 20x -H 84 = x» + 20X + 100 + 84 — 100 = (x +10)» —
(4)» « (x + 14) (X + 6).
9. x«4- 12x + 2 = x' -f 12x + 36 + 2 — 36 = (x + 6)'— 1/(34)«
= (x + 6 + V34) (x + 6— v/34).
10. x» + 3x + 5= x» + 3x + I -I-5— } = (x + |)' — x/"F^y?
= (x + 1 + >/— y ) (x + 1 - x/— y ).
11. x" — 10x + 24 = x« — 10x4-25+24 — 25 = (x — 5)«—(l)«
= (x — 4) (x — 6).
12. x» — 8x + 12 = x"— 8x + 16 + 12 — 16 = (x — 4)«— (2)»
= (x — 2)(x — 6).
13. x" — lOx — 24 = x»— lOx + 25 — 24 — 25 = (x — 5)« —
(7y = (x + 2)(x— 12).
14. X* — 10x» — 24 = X* — 10x« + 25 — 24 — 25 = (x* — 5)« —
(7)» = (x» + 2)(x» — 12).
15. x« + 4x' — 12 = x" + 4x» + 4 — 12 — 4 = (x» + 2)*— (4)»
= (x» + 6)(x' — 2).
16. a« + 2ab — U^ = a« + 2ah + 1/ — W = (a + i)'— (26)" =
(a + 36) (a — 6).
17. a» + 2a6 — tn" — 26m = a« + 2a6 + 6' — m* — 26m — 6^ =
(a + 6)'— (m + 6)' = (a + m + 26) (a — ni),
18. a« + 2a6 — m" + 26m = a« + 2a6 + 6"— m" + 26m — 6« =
(a + 6)»— (m — by = (a + m) {a — m + 26).
4
36 DIVISION.
DIVISION.
56. Division consists in finding how many times one quantity b
contained in another.
The quantity divided is called the dividend, that by which it is
divided; the divisor, and the result obtained the quotient,
57. It follows from this that the quotient multiplied by the divisor
must give the dividend.
The quotient is said to be exact when the dividend contains the
divisor an exact number of times. When this is not so^ the quotient is
called imperfect.
It will be seen that the object of division is to find a quantity called
the quotient; which; when multiplied by the divisor, will give the
dividend.
The result of the division of 4a;2; by 2a is plainly 2x, because 2a X
2x = 4ax; the dividend.
7a'c
So, = 7a, because 7a X ac = 7a'c.
ac
**~~ ^QX
But — gives — 2x for a quotient, because 2a X ( — 2x) a=s — 4ax,
2a
And — ^-^^^ = — 7a, because — 7a x ac = — *Ja^c,
la^c
ac
58. And, in general, since the quotient into the divisor must give .the
dividend ; when the sign of the quantity to be divided is unlike that by
which it is divided, the result will be negative, and when the sign of
the quantity to be divided is like that by which it is divided, the result
will have the positive sign.
Thus, =- s= + a, because + aX — 5 = — ab,
— 6
59. In Division, then, as in multiplication, like signs produce +,
and unlike signs — .
There are three cases in Division.
CASE I
When the dividend and divisor are both monomials,
60. Divide a* by a ; we are to find a quantity, which, multiplied by
o, will give a*. This quantity is plainly a', because a^ X a =^ a".
DIYIBIOK. 89
And, since, in multiplication, we add the exponents of like letters, we
mnst, in division, subtract tbe exponents of like letters.
Divide 6a' by 2a, the result is obviously 3o' ; because 2a X 3a" =
Co*. And since, in multiplication, we multiply the coefficients together
for a new coefficient, we must, in division, divide the coefficients for a
now coefficient.
Divide a'6 by a, the result is plainly ahy because ah X a = a^h.
Divide a*6V by a, the result is ah^c^. And, in general, if there arc
letters in the dividend not common to the divisor, they will enter into
the quotient with their primitive exponents.
RULE.
61. Divide the coefficient of the dividend by the cxtefficient of the
divisor for the coefficient of the quotient. Write after this new coeffi^
dent all the letters common to dividend and divisor affected with expo-
nents equal to the excess of the exponents of the dividend over those of
the same letters in divisor, and xcrite the letters common to the dividend
only with their primitive exponents. Give the quotient the sign -f,
where the monomials have like signs^ and the sign — , where they have
unlike signs.
EXAMPLES.
1. Divide x^y by x.
2. Divide — a^y by x.
3. Divide — T^yhj — x.
4. Divide x?y by — x.
Ans. a^y.
Ans. — x*y.
Ans. x}y.
Ans. — 7?y.
5. Divide 4a"Z»Pc by 2c*.
6. Divide — 40x"yV by
— 20x»^"'2;'»'.
Ans. 2a»iPc~».
Ans. 2x'-''f-^:n.
7. Divide xy'^ by x y.
Ans. X y.
3. Divide x^y by xy^.
9. Divide a?i^ by x*y^.
10. Divide xy by xV-
11. Divide 7?^ by «V.
Ans. tT y"^.
Ans. x^^yK
Ans. xy.
Ans. z'^s.
12. Divide 7?^ by aV.
13. Divide 1000a"6" by 500a-^6-.
Ans. zs"^.
Ans. 2a^h^.
14. Divide 1000a-"Z»- by
1 •
— 500a"6'.
Ans. —2ar^h'^.
I
1
16. Divide x ■ y" by x'y".
•
-L-a
Ans. X ■ y*.
i
i
40 DIVISION.
C2. It will be seen that the divicrion <^ monomials is impossible when
the coefficient of the dividend is not divisible by that of the divisor,
and when the divisor contains one or more letters than the dividend.
63. In such cases, the quotient appears in thd form of a fraction,
which may admit of farther reduction by striking out the common fac*
tors. Thus, the quotient arising from the division of 7a' by 2a"6, is
7a" 7a"
n~iTy because ^-^ X 2a"6 is plainly equal to the dividend 7a". But
7a« 7 . .
5-5^ may be reduced to — by striking out the common factor, a".
In like manner, 7a"6, divided by 2al^, is ^r-z- = --=- .
64. "We will now demonstrate two principles which will enable us
to reduce such expressions still lower.
1st. Any quantity raised to the zero power is equal to unity, that Is,
a**
a* = 1. For — by the rule for the exponents in division is equal to
a*
a"~" =B a«, but any quantity divided by itself is also equal to one.
Hence, — =: a« is also equal to 1. Therefore, a" = 1.
a"
In like manner, 2" = 1, and . (1000)« = 1, and (a + hy t= 1, &c.
65. 2d. A quantity may be transferred from the numerator to the
denominator, or from the denominator to the numerator, by changing
the sign of its exponent. For a~" may be multiplied by — ^ 1,
without altering its value.
-_ a" a* 1
Hence, a''^=sa~'^ X — a= — ss —
a" a" a"
But we know that a~»i=— -— , which we have seen, is also equal to — .
1 a"
The quantity a"™ has then gone from the numerator to the denominator,
by changing the sign of its exponent.
a*""
In like manner, a"*"" may be multiplied by -3- =» 1, without altering
its value.
„ a» ^ a~" «• 1
Hence, a" ==-=-== a" X -^ «= -— ^
1 a~* a~" or
m 11 Z"* ST' Z~9
So, l = ±xf_ = f_-l_=z-»,
zt z* 8r» z" 1
BIYISION. 41
The quantity 2' has passed from the denominator to the numerator
by changing the sign of its exponent.
1 1 zf z^ zv
So, — = — X— *= — = -r = 2»*
' z"' sr^ z' z"" 1
66. By the first principle, the quotient in Example 15 may be
-L-« -L-2
changed into x ° . 1 = x ■
By the second principle, the quotient of lo^h by 2a'& may be
7
changed into -h-2»^S aiid the quotient of 7a'6 by 2a J* may be changed
7
into -^aJr^,
Divide ic'y by a?t^c. Ans. x~*^""*c~', or
67. In this example, and in all similar examples, when the divisor
contains a letter or letters not contained in the dividend, we may con-
ceive this letter, or these letters, also to enter into dividend raised to
the zero power, and to execute the division, we have only to subtract
the exponents of like letters, as before. Thus, a^i/ may be written
a^C^f and we have only to subtract 1, the exponent of c in the divisor,
from o, the exponent of c in the dividend. The result will be — 1,
or — 1.
Divide oi?y by a'fcV ; then x'y =s a^j/a'*h'*c^, and the result will be
68. Strictly speaking, then, there is but one case in which the divi-
sion of monomials will not give an entire quotient, and that is, when
the coefficient of the dividend is not exactly divisible by the coefficient
of the divisor.
CASE n.
69. When the dividend is a compound quantity, and the divisor a
monomial.
The dividend may be regarded as the product of each term of the
quotient sought by the monomial divisor ; hence, to find this quotient,
wc must divide each term of the dividend by the divisor.
4*
42 DIYIBION.
RULE.
70. Divide each term of the dividend separately hy each term of the
divisor^ as in Case I., and connect the partial quotients hy their ap^
propria te signs.
EXAMPLES.
1. Divide 6xy — ^yz + 8a« by 2z,
Ans, SxV'^' — 2xV+4a.
2. Divide a? + 2ax + a* by x. Ans, a: + 2a + a'x'"\
8. Divide tp + x^^ + x»^ + x»^ by x».
Ans, 1 4- 35 + a?" + re*
4. Divide arp + ar«^» -f ar^^ + a;-»^ by or^,
Ans. 1 + ic -f a:* + a^.
5. Divide x""^"* — x~p~* — x-p-* — x""^"^ by x'.
^wfi. x-^P-' — x-^» — x-*»^ — x-^^.
6. Divide x~p~' — x-^^ — x-p-^ — x-P"^ by x~p.
Ans. x-' — x-^ — x-^ — tT'.
3 \
7. Divide 3x* + 2y -f «" by 2z. Ans. -^x'ar' + yar* +--.
Q ^
8. Divide 3x» + 2y + a* by 2ar'. ^n«. -^a^z + yz + -5-.%
9. Divide 40a»i»« — iPaftx + l^ahxy by 5a6.
u4n«. 8a6 — 2x + 3xy.
10. Divide 40a«6 — lOa&x + Ibabxy by — bab.
Ans. — 8a 4" 2x — 3a:y.
CASE m.
71. Wben the dividend and divisor are both polynomials.
We must keep in view that the dividend is the collection, after addi-
tion and subtraction, of the partial products arising from multiplying each
term of the quotient, when found, by each term of the divisor. If, then,
we can find a true term of the quotient and multiply it into each term
of the divisor, we will form so much of the dividend as was composed
of the product of this term by the whole divisor. And when we have
subtracted this product from the dividend, the remainder will be made
up of the partial products of the remaining terms of the quotient not
yet found by each term of the divisor. Now, if we can find another
DIVISION.
48
trae term of the quotient and multiplj it into the divisor, the prodact
will bo so much of the dividend as is made up by the multiplication of
this second term of the quotient by each term of the divisor. And if
we subtract this product from the first remainder, the new remainder,
if there be any, will be so much of the dividend as is made up of the
product of the remaining terms of the quotient not yet found by
each term of the divisor. We can thus regard each remainder in
succession as a new dividend until all the terms of the quotient are
found.
72. The whole difficulty then consists in finding, in succession, true
terms of the quotient. Now, we have seen that in the product of poly-
nomials there are always two terms which are dissimilar from the other
terms, and, consequently, irreducible with them. They are the terms
arising from the multiplication of the two terms in the multiplicand and
multiplier affected with the highest and lowest exponent of the same
letter. If, then, we divide that term of the dividend which contains
the highest exponent of a certain letter by that term of the divisor
which contains the highest exponent of the same letter, we are sure of
getting a true term of the quotient. For this term of the dividend has
been formed without reduction from the multiplication of the corres-
ponding term in the divisor by the quotient found.
73. In like manner, if we divide the terms affected with the lowest
exponent of the same letter, the one by the other, we are sure of get-
ting a true term of the quotient sought. For this reason, the dividend
and divisor are arranged with reference to the highest or lowest
exponent of the same letter, generally with reference to the highest.
When 80 arranged, the first term of each successive remainder will con-
tain the highest or lowest exponent of the letter according to which the
polynomials are arranged, and will, when divided by the fir<!t term of
the divisor, give a true term of the quotient.
The divisor in Algebra is written on the right of the dividend, and
the quotient immediately under the divisor.
Divide a* + 2ah H- 6" by a + 6.
Dioidaid. Divisor.
a« 4- 2ah -f h^
a -{- h
a" -f ah
a + h
-i- ah -f h^
ah + y
Quotient,
Dividend.
or h^ -f 2ha + a^
y + ha
ha + a'
ha + a"
Divisor,
h +
a
h -^ a
Quotient,
-f- Remainder,
-f- Remainder.
44 DIVISION.
74. We see, that by arraDging dividend and divisor with reference
to the ascending or descending powers of the same letter, and dividing
the first term of the dividend and the first of the remainder by the first
term of the divisor, we necessarily get true terms of the quotient. Btit
if the polynomials are not arranged we will not get true terms of the
quotient. Write the dividend and divisor thus,
Dividend,
6« + 2ah + a«
Divisor.
a + 2»
a
— — &c., Quotient
a
a
We get — for the first term of the quotient, which is not a true
result. It is also plain that the division would never end.
75. Since each successive remainder may be regarded as a new poly-
nomial, to be divided by the divisor, we can change the arrangement
of the remainders at pleasure, provided we make a corresponding change
in the divisor. In the example above we might arrange the remainder
ah + 6' with reference to &, and write it &* + aft, provided we change
the arrangement of the divisor, and write it h + a.
RULE.
76, Arrange the dividend and divisor with reference to the oMendr
ing or descending powers of the same letter ^ and then divide the first
term on the left of the dividend hy the first term on the left of the divi-
sor. This will give the first term of the quotient.
Multiply this term into eaxih term of the divisor ^ and subtract the
product from the dividend. Divide the first term of the remainder by
the first term of the divisor for the second term of the quotient. JHulti'
ply this term into the divisor, as before, and subtract tJie product from
the first remainder.
Continue this process, dividing the first term of each remainder by
the first term of the divisor until we get a remainder, zero, when the
division is said to be exact.
But, if the eocponents of the letter, according to which the arrange-
ment is made, are all positive, and the first term of any remainder %$
DIVISION. 46
not divisible hy the first term of the divuor^ the quotient is incomplete;
OTy t/some of the exponents are negative^ and the Utter, according to
which the arrangement has been made, has disappeared from any of
the successive remainders, the qitotient is also incomplete.
EXAMPLES.
1. Divide a^ + 3ax* + 3o*x + a* by x + a.
Ans, x' + 2ax + a'.
2. Divide x* + ^aa? + 6a'x* + 4a'x + x* by x + a.
Ans, X* + Sox* + 3a'x -f a*.
3. Divide x* — 3x*y + 3xy' — y* by x — y.
Ans, x" — 2xy + y".
4. Divide 4a« — 6« by 2a + 6.
-4»?<. 2a — 6.
5. Divide x" — y" by x — y.
^7w. x»-* + x»-V + a— y + a ^^ — sT
X — y
6. Divide of — y*hj oc^ — y*.
Ans. x' + xy -f xy + y*.
7. Divide x" — y" by x* — y«.
^n«. x'« 4- xy + x'y* + xy + xy + y'^
8. Divide x' — y' by x' — y*.
-4n<. x" + xy + xy* + xy* — y^
a^-y'.
9. Divide x' - — y® by x* — y*.
-4n». X* + xy + y .
10. Divide x? + y" by x« + y*.
Ans. x' — xy + xy — y* + 2y'
11. Divide x« + y« by x« + y».
Ans. x'° — xy + xy — xy + xy — y^*^ + 2y*»
aj* + y
12. Divide x' + y' by x« + y».
Ans. X* — x'y' + xy* — xy* + y'
i?Ty
13. Divide x* + y* by x* + y.
Ans. x^ — xy + y*-
46 DIVISION.
14. Divide x** -h y'* by a;* + y*.
16. Divide a;»* + y» by a;» + y*.
^n«. a:" — a^ + a^V — a^ + y*".
16. Divide lax — lay + 7a6 + hx — hy + l^ by 7a + 6.
Ant. X — y + ft-
17. Divide xy^ + x * + ar'y" + y"* by x" + y".
Ant, x~* + y^.
18. Divide 14x' — 14y* by 7x« — 7y".
Am, 2x« + 2xy + 2x»/ + 2y*.
19. Divide a* — 2ax + x" + 2ay — 2xy by 2y + a — x.
j4n«. a — X.
20. Divide ax«— 2arx+ar*-f 6x"— 2?>rx+«'^+ci^— 2rrx4-cx«by.
Ans, (x — r) (a + i 4" c)-
21. Divide «* — 2«x + x» + r« — 2xr + r" + r« by « — x + r.
Ans. 8 — X + r.
22. Divide 50x* + 50y» by 25x» + 25^.
Ans. 2a* — 2xy + 2y«.
23. Divide (« + 6 + c) (x — r)» by (a + 6 + c) (x — r).
j4n«. X — r.
24. Divide x^y-* + a^b + a^ + y + y*b + ar^y* by x* + y*.
Ans. y~^ + h + or*.
25. Divide 49xy — 64xy by 7x'y* + 8xy.
Ans. la^y* — 8xy.
26. Divide a« — &• by a» + V.
Ans. a* — a«6« + b* — 26^
a» + t».
27. Divide a'« — 6" by a« + ft«.
jln«. a" — a«6' + a*2>* — a^b^ + b^ — 26'°
a« + bK
28. Divide a' — 6' by a' + 6«.
Ans. a« — flW + a»6* — 6«
29. Divide a»' — 6'« by a» + 6».
-4n«. fi» — aV + a^'b'' — 6».
30. Divide a" — 6" by o» + 6*.
^fw. o» — a'°6» + a*6~ — 6»
31. Divide 4 by 1 + x.
Ans. 4 — 4x + 4x* — 4x' + 4x*
"IT*.
DIVISION. 47
5 32. Divide 4ljx + 1.
Ans. 4ar» — 4ar^ + 4x-* — 4ar*
X + 1.
O 33. Divide m — nx + pa^ — yx* by 1 + x.
Ans, m — (m + n) X + (m + « + 1>) x" — other terms.
34. Divide 1 + x by 1 — x.
Ans, 1 + 2x + 2x' + 2x» + &c.
35. Divide 1 — x by 1 + x.
Ans, 1 — 2x + 2x* — 2x» + &c.
36. Divide x' + ay + ax + 1 + xy by x + a.
Ans, X + y + 1
X + a.
37. Divide x» + x« + 5 by x« + 2.
Ans. X + 1 + 3 — 2x
x^ -f- 2.
Remarks.
77. I. When the exponents of the arranged letter in dividend and
divisor are all positive, the division will not be exact if the first term
of any of the remainders does not contain the arranged letter to a higher
power than it is contained in the divisor. When, therefore, we get a
remainder in which the first term contains the arranged letter to a lower
degree than it is contained in the divisor, we need proceed no further,
for we can never arrive at an exact quotient.
If some of the exponents of the letter, according to which the poly-
nomials are arranged, are negative, the last rule will not hold good, for
the result of the division will be a term affected with a negative expo-
nent, and there ought to be one or more such terms in the quotient.
If, however, the letter, according to which the arrangement has been
made, has disappeared from any of the successive remainders, we need
proceed no further.
n. Examples 6, 7, and 9, show that the difference between the like
powers of two quantities is divisible by the like powers of a lower degree
3f those quantities, when the common exponent of the dividend, divided
by the common exponent of the divisor, gives an exact quotient. And
they show that the number of terms in the quotient is expressed by the
quotient of the exponents. Thus, in example 6 there are four terms
in the quotient, the same as— the quotient of the exponents. In ex-
48 DIVISION.
ample 9, the quotient of the exponents is three, and the division is
exact, with three terms in the result.
m. Examples 13, 14, and 15, show that the sum of the like powers
of two quantities is divisible by the sum of the like powers of the same
quantities when the result of the division of the exponents is odd.
lY. Examples 28, 29, and 80, show that the difference between the
like powers of two quantities is divisible by the sum of their like powers
when the quotient of the exponents is even. In the last two cases, as
in the first, the quotient of the exponents gives the number of terms in
the result.
V. If all the exponents of the dividend and divisor are positive, we
can tell, by a simple inspection, in many examples, whether the divi-
sion is possible; when the extreme terms of the arranged polynomials
are not divisible by each other, the quotient will not be exact. Take
Q^ -f 2xy + y", to be divided by x' + ^/r' ; the division is impossible,
because the exponents of the result must be positive ; and o^ by o? will
give re"*. It, of course, does not follow that the division will give a
complete quotient when the extreme terms are divisible by each other.
Take a?" + 4 H- y*, to be divided by a; -h y ; the extreme terms are
divisible by each other, but the quotient of the polynomials is not
exact.
PRINCIPLES IN DIVISION.
78. I. We will now show that the difference between the like powers
of two quantities will be divisible by the difference of their like powers
of a lower degree, whenever the quotient of the common exponent of
the dividend, by the common exponent of the divisor, is exact. That
is, that a" — 5" is divisible by a" — 5°, when m is exactly divisible
by n.
a" — 5" I a" — i»"
(x« — a"-"ft" a'
a'"6" — i" == 6" (a"-" — 6"-").
Performing the division, we get a"*^ for a quotient, and &" (a"*~' —
6"~*) for a remainder. If this remainder is divisible by a" — i", the
dividend will also be divisible by a" — 6" ; for, represent the dividend
by D, the divisor by (7, the quotient by 9, and the remainder by R.
Then a" — 6", or D s=s jrf + R. Now, qd is plainly divisible by f/, and
if R be also divisible by cf, the first member must also be divisible by
dy otherwise we would have the sum of two entire quantities equal to a
y
DIVISION. 49
fraction. Hencey in general, if the remainder is divisible by the divi-
sor, the dividend will also be divisible by it. Then, if we can prove
that the remainder, h' (a"~" — ^"~")> can be divided by o" — h% we
can prove that the dividend can also be divided by a" — I'. But if the
factor, a"~" — 6"~", is divisible by the divisor, 6" times that factor will
also be divisible by the divisor ; and the remainder giving an exact quo-
tient, the dividend will also give an exact quotient. That is, a" — b^
will be divisible by a" — 6° whenever a""" — Z»"~" is divisible by a" —
&■; or, in other words, if the difference of the like power of two quanti-
ties is divisible by the difference of the like power of the n** degree,
the difference of the like powers of a degree higher by n will also be
divisible by the difference of the n*"* degree. But we know that a" —
l^ is divisible by itself, a" — fe" ; hence, by the principle just demon-
strated, a^ — 6^ must be divisible by a" — Z»". And since o** — 6** is
divisible by a" — I/*, a** — ft*" must also be divisible by a" — i", and
so on, for powers of a degree greater by n, until it finally reaches and
divides a" — 6"*. This power must eventually be reached, because m
is supposed to be a multiple of n.
79. In Example 8, x' — y, divided by x* — y', the quotient was
not exact, because m or 7 was not a multiple of n or 2.
In Examples 6, 7, and 9, the quotients were exact, because in each
case m was a multiple of n.
80. It is plain that pa" — joZ>" can be divided by ja° — qh^ when p
and m are multiples of q and n. For we can put the expressions
under the form of p (a" — i"), and q (a" — •' i"), and the quotient
will, of course, be exact when p will divide q, and a" — J" will divide
a" — b\
Thus, in Example 18 we found 14x' — 14y' divisible by 7x' — 7^ :
the dividend can be written, 14 (as* — ^, and the divisor 7 (a^ — y*),
and since 14 will divide 7, and x^ — y" will divide cc* — y*, the quo-
tient is exact.
81. The demonstration holds good whenever ti, added to itself a cer-
tain number of times, will produce m, and is therefore true for quanti-
ties affected with negative and fractional exponents.
«
Divide ar^ — h^^ by a'^ — b'\ Ans. a"* + ft-".
Divide a-* — ft"* by a"* — ft-*. Ans, a"* + ft"*.
Divide a"* — ft-« by a"* — ft-'. Ap^, ar^ -f a-"ft-«-f ft"^.
60 DIYISION
Divide a* — h by a — 6* Am. a + h*.
Divide J — h^ by a* — h*. Am. a + ah^ + h.
Divide a^ — h^ by a* — 6*.
^iw. a* + a*6* + a V + a*6* + ft*
82. n. The difference of tbe like powers of two quantities is divisi-
ole by the sum of their like powers of a lower degree, when the result
pf the division of the common exponent of the dividend by the common
exponent of the divisor is an even number ; which is evident from the
principle of Article 50.
83. lU. The sum of the like powers of two quantities is divisible
by the sum of their like powers of a lower degreCi when the quotient
arising from dividing the common exponent of dividend by the common
exponent of the divisor is an odd number.
That is, a" + 6" is divisible by a" + i", when m, divided by n, is
an odd number.
Performing the division, we will have — 6" (a"~* — 5""") for a re-
mainder, and whenever the factor, a"~" — &"*"", is divisible by a* + 6",
the dividend a" -f 6" will be divisible by the -same divisor. But we
have just shown that a""" — ft""" can be divided by a" + 6" when
IS an even number, and since is the same as 1,
n ' n » '
the quotient of — must be odd in case the quotient of is even.
n n
Hence, the truth of the proposition. Thus, in Examples 13, 14, and
15, the quotients were exact, because for each of these examples —
n
was an odd number. So, in like manner, oc^ •^- a', re* + a*, x' + a',
&c., are divisible by x + a, and give 3, 5, 7, &c., terms in the quo-
tient, with signs alternately plus and minus.
84. lY. It is plain that the sum of the like powers of two quantities
cannot b^ divided by the difference of their like powers; that is, a" -h
6" cannot be divided by a" — b*, because it is not possible to deoom-
p6se the sum of two quantities into factors, one of which will be a dif-
ference between the quantities.
AI.OXBKAIO FBAOTIONB. 51
ALGEBRAIC FRACTIONS.
85. A firaetion is a broken part of unity. The denominator denotes
the numher of equal parts into which the unit has been broken or
divided^ and the numerator expresses the number of these equal parts
taken. Thus^ the fraction } indicates that the unit has been divided
into three equal parts, and that two of these parts have been taken. In
like manner^ the fraction -7- indicates that the unit has been divided
into h equal parts, and that a of these parts have been taken.
86. Eveiy quantity not expressed under a fractional form is called
an entire quantity.
87. An expresssion, made up in part of an entire quantity, and in
part of a fraction, is called a mixed quantity. Thus, 4 + i^, and a —
— axe mixed quantities.
88. A proper fiaction is one in which the numerator is less than the
denominator. An improper fraction is one in which the numerator is
greater than the denominator.
89. A simple fraction is one whose numerator and denominator are
simple quantities. Thus, -7- is a simple fraction.
90. A compound fraction is one which has a compound expression
in the numerator or denominator, or in both. Thus, , :,x
and z are compound fractions.
91. The minus sign before a fraction changes the signs of all the
terms in the numerator. Thus, r — is equivalent to — ; — .
6 6
92. A few of the principles involved in operations upon fWictions
will now be demonstrated.
I. The multiplication of a fraction by an entire quantity is effected
by multiplying the numerator, or dividing the denominator by the
entire quantity.
For to multiply -r- by c, is to repeat -r-, c times, and since -j- taken
o o
52 ALGEBRAIC FRACTIONS.
twice^ is -j--y three times is — , &c, the result of multiplying y , c times
etc
^ will plainly ba — . The multiplication can also be performed by divid-
ing the denominator by c; for, if we divide the denominator by c, and
^ is the quotient of the division, then the unit will be divided into e
times fewer parts than before, and, of course, each part is c times as
great as before. Now, if we write a over q, we will have taken as
many parts of the unit as before ; and since each part is c times greater,
the fraction — is plainly c times greater than the fraction -j- .
The multiplication of the numerator by a whole number may also be
demonstrated in another way. -7- is taken to be increaFcd c fold, and
if the denominator remains unchanged, while the numerator is multi-
plied by r, the number of parts into which the unit is divided remains
the same, but the number of those parts taken is increased c fold, the
ttC
new fraction y must then be c fold greater than the old.
93. II. The division of a fraction by an entire quantity is effected
by dividing the numerator, or multiplying the denominator by the
entire quantity.
For, to divide -7- by c, is to diminish the value of the fraction c fold,
and if the result of the division of a by c is g' ; then 5 is c times smaller
than a J and y- is c times smaller than -7-. Because, while the parts of
the unit have remained unchanged, c fold fewer of these parts have
been taken.
The division can also be effected by multiplying the denominator by
r, for then the parts into which the unit is divided being increased c
fold, the value of each part must be decreased c fold, and, of course,
when the same number of these c fold diminished parts are taken, as at
first, the result must be c times smaller than before.
94. III. The value of a fraction is not altered by multiplying the
numerator and denominator by the same quantity. The fraction -7- is
not altered in value by multiplying the numerator and denominator by
c. For, to multiply the numerator by c is, from what has just been
ALGEBBAIO FBACTIOMS. 58
shown, to increase the yalae of the fraction c fold; and to multiply the
denominator by c is to diminish the value of the fraction c fold, and,
of course, the fraction has undergone no change of value since it has
been increased aod decreased equally.
95. IV. The value of a fraction is not changed by dividing the
numerator and denominator by the same quantity. Because the two
divisions cancel each other, and leave the fraction in its primitive
condition.
96. .V. If the same quantity be added to the numerator and denomi-
nator of a proper fraction, the value of the fraction will be increased.
To show this, it will be necessary to show that if the same quantity
be added to two quantities differing in magnitude, the smaller of these
will be increased more, proportionally, than the larger. Take the num-
bers 1 and 2, add 1 to both, the first will be doubled, but the second
will not be ; add 3 to both, the first will be increased four fold, while
the second will only be 2} times greater than before. In general, let
a and b represent the two quantities, a being less. than b; add a to both|
the first will be doubled, but the second will not be, because a + ^ is
less than 2b. Now, when the same quantity is added to the numerator
and denominator of a proper fraction, the numerator is increased more
proportionally than the denominator ; the number of parts taken, then,
is increased without their being an equal decrease in the size of those
parts. Hence, the new fraction must be greater than the old.
97. YI. By adding large quantities to the numerator and denomi-
nator, we can make the value of the fraction approximate indefinitely
near, though it can never become unity. Thus, add 1000 to the nume-
rator and denominator of i, the new fraction is -j-JJ^, almost, though
not quite, unity. Now, add a million to both terms of the fraction, and
the result will be indefinitely near, without being altogether unity.
98. Vn. If the same quantity is added to the numerator and deno-
minator of an improper fraction, the value of the fraction will be
decreased.
For, the denominator being smaller than the numerator, will be
increased more proportionally, and the value of the fraction must be
diminished.
Take | as an example, add 10, and the fraction becomes || <[| |.
It is plain that no addition to the numerator and denominator can ever
reduce the fraction as low as unity.
5*
64 ALOXBRAIC tRACTIONS.
REDUCTION OF FRACTIONS.
99. The reduction of fractions consistfi in simplifying their fonns
thout altering their
Their are 10 cases.
without altering their value.
CASE I.
To reduce a simple fraction to its simplest form.
RULE.
Strike out all the f acton common to the numerator and denomi-
nator y and the result will he the fraction reduced to its lowest terms.
Take as an example, :
, We know the factors of a* — 6" to he (a 4- 6) (a — h), hence,
a^ — y" (a — 6) (a + ft)
— --=- = -^^ <-^^ == a — 6.
a -H 6 + 6
The expression a + 6^ heing common to numerator and denominator,
may be stricken out, since, by Article 95, we can divide both terms of
the fraction by a + & without altering its value
Take "I+l«+i.
n -h 1
We know, by Article 48, that n' + 2n -f 1 =: (n + 1)', hence,
n« -h 2» + 1 _ (n -f 1) (n + 1) _
n -f 1 - (n + 1) — « + A-
ft. Reduce —
4. Reduce
5. Reduce
X — a Ans, x — a,
12a*— 12h'
4a — 46 ■ Ans. 8 (a + 6).
6a*occ + 6axm Ann. nc -{- m
36ox -f 42aV 6 H-7oV '
ALdXBBAXO VRAOTIOMS. AS
^ ^ , 4m" — 4»* . w — «
6. Bednoe s-t — : — ^ -^nt. — 5 — .
7. Beduce
8. Reduce
8 (m + «)
ac -f- ftc + c^ + (/c
a H- 6 +c
+ cf •
3a;y +9:^*/-
-27oi?^d
Afu, c.
96xy H- ^axy + 12xy '
. 1 + Sxy — 9xyc
' 36 + a + 4xy
CASE n.
100. To reduce a mixed quantity to the form of a fraction.
RULE.
Multiply the entire quantify by the denominator of the fraction;
connect the product with the numerator by the appropriate sign of the
numerator^ and write the denominator under the whole result.
X ab I 35
Take for an example, a -f y, the result is — r — . We have not
altered the value of the fraction, since the division by b will again give
X
ufl a + 7-. We have, in fact, only multiplied and divided a by the
same quantity, b.
X —— a* . * as
2. Beduce a H • Ans. — .
8. Reduce 2x + "^ -~_t__v;^^ ^^^ — +_o ^
5. Beduce
a+-
X~
2x
, -•
—
2ax -f
a»
r
a
•
m-
-n +
m'
— n«
m
H-«
a —
2x H-
a«
— 2ax
+
x^
X — a.
i
1 4
. r. M
2-
— a^
4. Beduce m — n H . Ans. 2 -.
m + n
Ans. — X.
3 — 2a:«
6. Beduce 1 4- x + = . Ans, ^ —
1 — X 1 — X
66 ALaBBBAIO FBA0TI0N8.
7. Reduce 7 - a + ^+i. Ans. ^HZ^i+^jL*.
8. lUduoe 4 ^^^ • Ans. -^-^.
/\ -Ti -I J ft ^* ~"~ 2(ix 4- 3a*
9. Keduco 4x + 3a .
X + a
9aa; + 3a;'
Ans,
X -^ a
10. Eeduce x — a H . Ana. .
a; 4- « X -\- a
11. Reduce 2> + 2 v + ^^^^^^ — ^ — ^ZZK,
+ y Anx. y.
io f> J -11 11 11 lly* — llx« + llx— 33y + 22
12. Reduce llx + 11 v — 11 H — = — ^^—^ — .
X — y + 1
Ans, 11.
CASE m.
101. To reduce a fraction to an entire or mixed quantity.
RULE.
Divide the numerator hy the denominator for the entire partj and
place the remainder j if any, over the denominator for the fractional
part,
a;* 4- a*
Take
X 4- tt
We have a right to divide both terms of the fraction (Article 95) by
the denominator^ and we will not alter the value of the fraction. The
reduction in Case III, is nothing more than dividing both terms of the
fraction by the denominator, and; since the new denominator is unity,
it need not be written.
_- a;" 4- «' a^ 4- <^' a; + a
Hence, — ; = — ; ; ; — ,
a;4-a x4-a x + a
that is, both terms divided hj x + a.
ALOBBBAIO FBA0TI0N8.
57
The result is a; +
— osB + a'
a? + a , or simply, x +
— ax + a*
X '\- a'
1. Reduce
2. Reduce
3. Reduce
4. Reduce
5. Reduce
6. Reduce
7. Reduce
8. Reduce
9. Reduce
10. Reduce
a' + jfc"
a
Am, a + — .
a
a^—2ax-\' a« + l
X — a
2 (w' — n*)
m + n
1 +x
1 — x
\+x
1
1
7a« + 76'
r
7a' + y
a' + 6«'
g* — 6'
a — 6'
-4n«. ic — a +
a — a.
-4n«. 2 (w — n).
. - 2a;
-4.IW. 1 +
An$. 1 —
An$, 1 —
-4.n«. 1 4- -= .
\ 1 — X
Ans 7.
A ^ 6^
Ans, a* + aft + 6^.
1 IB-
2x
l+x
X
l+x
x
CASE IV. .
102. To develop a fraction into a series.
RULE.
Divide the numerator hy the denominator, as above, and co7Uinue
the division as far as may he rehired, or may he deemed necessary.
68 ALOSBRAIO V&AOTIONS.
This case differs from the last in two particulars; Ist, It IncludeB
only fractions which cannot be reduced to entire quantities. 2d. The
division; instead of stopping with one term of the quotient; is carried
on as far as may be thought proper.
EXAMPLES.
1 + x
1. Expand = into a scries.
Ans. 1 -f 2x + 2x' + 2x* + &c., indefinitely.
1 X
2. Expand z into a series of five terms.
An$. l — 2x + 2a^ — 2a^ + 2a;*— &c.
8. Expand into a series of six terms.
Ans. 1 — X + X* — x^H- X* — a^ -|- &o
4. Expand ^ into a series.
X ^^~ X
Ans, 1 + X + x" + a^ + «* + &o.
7a« + h* .
5. Expand — r-r-« '^^^
a' + o*
a series.
, ^ 66* 6b* 6i« . ,
Ans, 7 5 H — ^ ^ +&C.
a
6. Expand =• into a series.
"^^ X "J" X
j4ii«. X""* X"* X~* X~^ X"* &0.
1
7. Expand — j— r iuto a series of five terms.
X ^" X
Ans, x~* — x^ + X"* — X"* + x-* — &o.
x"
8. Expand 75-=-- — ^ into a series.
X — Jx -|- of
Ans. X + 2x" + 3x' + 4x* + Sx" + &c.
9. Expand "!"^ .
.21^ 2y . 2y* 2«*
10. Expand
ALGBBBAIO TBAOTIONS.
4
Afu, 4x""* — 4a;~'y + 4x"y — 4a;^^ + &o.
103. In all tbese examples^ since the quotient is incomplete^ it is
plain that the prodact of the divisor by the series will not give the
dividend.
104. The different results of the same division in 4 and 6, 3 and 7|
show that the series may be changed by changing the order of the
terms in the divisor ; and it might readily be shown that a change in
the order of the terms in the numerator (when there is more than one
term) will produce a corresponding change in the series.
CASE V.
105. To reduce fractions having different denominators to equivalent
fractions having the same denominator.
RULE.
Multiply each numerator into all the denomtnatorSy except its oum,
for a new numeratory and all the denominators together for a common
denominator of all the new numerators; or, multiply each numerator
by (he quotient arising from the division of the least common multiple
of the denominators by each denominator respectively, and write the
products thus formed for new numerators over the least common multi-
ple as a common denominator.
X u
Redace — and — to a common denominator.
a c
ex ao
By the first rule we set — and — , and it is plain that these frac-
■^ ^ ac ac ^
tions have the same value as at first; since each has been multiplied and
divided by the same quantity.
X o
Reduce -5 and — to a C3n:mon denominator,
a* a
60 ALOEBBAIO FBA0TI0N8.
The least common multiple is a' ; the quotient of this common muld-
ple bj the first denominator is 1, and this is^ then, the multiplier of
.a'
the first numerator. The multiplier of the second numerator is — => a.
a
Hence, the equivalent fractions are -^ and -^ . It is plain that each
fraction has been multiplied and divided bj the same quantity.
3. Reduce -r-. -^i and — to equivalent fractions with a common
a* or a ^
denominator.
. X ay a*b
or a* a*
4. Reduce i, a, -j-, and -^ to equivalent fractions having a common
denominator.
. 26 4ab ha^ , 4y
^"'- W lb' iP "'^ 46-
5. Reduce }, ; — , -, ■ — r-., and --r to equivalent fractions with
a common denominator.
2 (m + n)« 4(m + w)'(m — n) 4 (x' + o") ^ (m -f «)*6
'• VCm -f n/' 4 (m + n)« ' 4 (m + nf 4 {m+rif
6. Reduce — , , -, to equivalent fractions having a common
denominator.
. a* ax* + a*x aa:*
•Am, —^ , u , — =-.
a-x a,x arx
7. Reduce , — , , — to equivalent fractions havine a
m — n b X c ^ °
common denominator.
bca? (w — n) c*x (m — nfbc (m — n) Vx
An$.
(m — n) 6cx' (m — n) bcx' (m — n) bcx^ {m — n) teas*
By reducing the above results by Case I., we will get back the ori-
ginal fractions. We can thus verify the correctness of the results
obtained.
ALOEBBAIC FBAOTIONS. 61
CASE VI.
106. To add firactional quantities toother.
RULE.
Reduce the fractions^ to a common denominator j if necessari/^ and
over this common denominator write the algebraic sum of the new
numerators.
Fractions, which have different denominators, cannot be added to-
gether previous to reduction to the same denominator, because they
represent different things. Thus, i and i neither make } nor | ; the
first indicates that one of the three parts into which the unit has been
been divided has been taken ; the second indicates that we have taken
one of the four parts into which unity has been divided.
Since, therefore, the parts taken are different in magnitude, they
cannot be added together. It would, obviously, be just as proper to
add a peck (the fourth of a bushel) and a quart (the thirty-second part
of a bushel), and call the sum two pecks, or two quarts, as to add two
fractions with different denominators.
EXAMPLES.
1. Add — H , and together. Ans, .
a a a ^ a
2. Add — , + — and . Ans. 3 .
a* a a cr
Q AJj^ + ^x^— * A 4c + cr — 26
o. Add — -r — to . Ans. ^ .
2 c 2c
4. Add X — a -^ ^ . Ans. —5- -— .
2x — Za Z (x — a)
c A jj 2a5 . 3x 4a; . 6a; . . 164a5
5. Add -g- + ^ - -5- + -g-. An,. 3^.
6. Add ^IHi? + -L_ to ^+J' - !!i.
X X H- y c n
. nc (x" — y") + hcxn + nx (x -f y)' — mcx (x + y)
ncx (x + y)
62 ALOSBRAIC FRACTIONS.
7. Add ^ + — to — --^ H .
X — y m X +y c
J 2mc ( x' +30 + (c' -f w*) (a;' — y^
mc {per — y)
8. Add 4 + 4x to J + -?-. ^ns. lli^j±i).
4 4
9. Add 4 — 4a; to I — -?-. Am. ILCljZf),
4 4
10. Add 4x — 4 to -^ — i. Ans, ^7 (x— 1)
4 4
. Add — r ; — to « — a +
b X + a X — a
(^x + a + hx — ha) (x» — a') + 2al^
^'^' bQ^ — a*) •
Remark.
107. When none of the terms have been reduced with each other,
the result divided by the common denominator ought to give back the
original series of fractions. This may be noticed in the first three
examples.
CASE vn.
108. To subtract one or more fractional quantities from one or more
fractional quantities.
RULE.
Make all the denominators the same, if the fractions have not a com'
mon denominator, then write into one sum the numerators of the minu-
end, and from this sum take the algebraic sum of the numerator, or
num.erators of the subtrahend.
Write the difference over the common denominator.
The same reasons which show that we cannot add fractions, with
different denominators, prove that we cannot subtract fractions with
unlike denominators. The fractions represent different things until
reduced to a common denominator.
AIOJBBAIO FBA0TIOK8.
KTiwpT.nn
1. Subtract ■§. from £
6'
2. Subtract ~ from ~
6'
• ^- ^'^"^ *e differeucc between a; and ^
2'
^n». f-or — £..
4. Take the difference between «+ ± „„d «__ « . ^
2 2'
^ns. +a,or — o.
5. Subtract ^^ + |. f„^ a--J ^
2 — '°' — (* + «).
6. Subtract ^'^j .^f a — b
^««. - 2 i^ 2x)
2 ' or — (5 + 2a;).
7. Subtract 2a: + I + 4 fr^ y+^
d 2 •
Ant. Hl+^t z}^"" ~ 2a — 24
8. Subtract^* fi„^2. + | + 4. '
6
10. Subtract i^Zl5 fron, « + * „ . .1
3 from -__. ^^ a + 5^
11. Subtract « ~ f. from ? + g
^ 7»
m6
64 ALGEBRAIC FRACTIONS.
CASE Ym. «
109. To multiply fractional quantities together.
RULE.
lfi}ie quantities are mixed , reduce them to a fractional form, then
multiply their numerators together for the numerator of the product
required, and their denominators together for the denominator of this
product.
a c
Let it be required to multiply j- by -7.
a a
To multiply -7- by (?, is to repeat -7-, c times ; the result will then
ac
plainly be -p. Because, while the size of the parts into which the unit
has been divided has remained the same, the number of parts taken has
been increased c fold. -^ is then, obviously, c times greater than-j-.
But we were not required to multiply ~ by r, but by the quotient ari-
sing from the division of c by d; our multiplier has then been d times
too great, and, of course, the product is d times too great. The result,
then, must be corrected by dividing by d, Ilence, the true product
of -r- by -7 is — ,, in accordance with the rule.
b ^ d bd'
EXAMPLES.
1. Multiply 2x + -^ by -r-. Ans,
2. Multiply "-^ by ^^'. An
8. Multiply 1- by 1-±JL. A
c a
4. Multiply "''■^ ^ by '^~^ . Ans. . ,
a;+y » — If ar — y
4
•
jt. g ,
ae
ALGEBRAIC FEAOTIONS. 66
.-t j^ -.-a /*-! «-»
6. Multiply ^L+JL.hj '^ y" .
^"'- (a + 6) (X- V y-')
6. Multiply^ by ^. ^^,,
7. Multiply 7 + ,=3^ byt^^. ^m». -.
7— y I If + IT y
8. Multiply c + - by - + ^. ^n*. ^l±i^'.
y c c« xy
9. Multiply LLliZiy byl=i^.
1-160;' + 8xy-y '
10. Multiply ___^by3— 3^-p^.
*• 1 — 16x« + 8xy — y«'
11. Multiply S+o- — cbya — y + T*
a4x + 4ax — 8ra; — 32a — 2a» + bac + 16c — 2c«
Am,
8
12. Multiply |- — J ^y |- + J-
^^- — 36— ^'4-9-
CASE IZ.
110. To divide fractional quantities by each other.
RULE.
If the quantity to he divided is not in the form 0/ a simple fraction^
reduce it to that form; and if the quantity to he u$ed gw a divisor is
not already a simple fraction, make it so. Then invert the terms of
the divisor and proceed as in tlie last case,
6* B ^
66 ALQEBRAIO FBACTIONS.
CL C
Take, as an example, -y- to be divided by -7.
The result of the division of -r by c must be c times smaller than-r-
It will then plainly be j-, because, while the number of parts taken has
remained unaltered, the size of these parts has been diminished c fold,
since their number has been increased c fold. But we were not re-
quired to divide -j- by c, but by the quotient of c by d. Our divisor,
then, has been d times too large, and the result, j-y of course, d times
too small. The error, then, must be corrected by multiplying by d.
Hence, — -f- -r = r-j ^^ accordance with the rule.
6 d be
111. The demonstration is general, and is applicable when the divi-
dend and divisor are mixed quantities, or made up of several fractions ;
or, when either dividend or divisor is a mixed quantity, or composed of
more than one fraction. For the fractions, if not already under the
form of -J- and — , can be put under these forms without difficulty.
The demonstration in the last case is general, for the same reasons.
EXAMPLES.
1. Divide ^ by -5-.
Ans. |x.
2. Divide -^- by 6^ Ans, zr^,
5 -^ 26
3. Divide -=- by 4. . .
^ 1 X + 1
4. Divide — 3- — by — 5 — . . . ,
4-^8 Am. 2 (a; — 1).
5. Divide a = — by = . . -
or — 1 X — 1 Ans. 1.
6. Divide . ^ by -5 jj. a ^ ^ ^ xn
a + o •'a* — tr Ans. (x — y) (a — b).
ALGSBSAIO. FBACTIONS. 67
7. Diyide m by ^= ^ y. Am, *"^
8. Divide ^ by ^^^-=^ + 2x, Am. "^^
11. Divide -i^ — -^ 6 + cc by t-t — ^\
9. Divide 1 4- 7> 4r by — 1 — -^ + 4x.
o " 6 Aim, — 1.
10. Divide 1 + ^— '4xby + 1— |' + 4x.
3 + rc»— 12x
(aj — y) a 4- a'x
Am. 2 1^",-^ ^ ^" ^ ^)l'.
a*|a5 — y 4- ax
12. Divide 2(x+y)+a5_ 4(x + y)+2ax
a •' (x — yja + arx
to Tn -J. 2a + X + 2 as , »» + n a;
13. IhTide .J 2c + jhj-^ J.
m 4- » — 2Da;
112. The quotient, multiplied by the divisor, ought to give the divi-
dend ; and in the last case, the product, divided by one of the factors^
ought to give the other factor. We are thus enabled to verify our
results.
We have seen that both the numerator and denominator of a fraction
can be multiplied by the same quantity without altering the value of
the fraction. Hence, by multiplying the term of a fraction by minus
X 1
unity, the signs of these terms may be altered. Thus, — 7 — may be
- ^ c
written ; — -^may be written — ^, &c. Hence, — j =3 —-.
That is, the quotient of two negative quantities is positive.
So, 7, or, — =; =: — -=-, read minus the fraction -=-.
+ o — 00 o
68 ALGEBRAIC FRACTIONS.
CASE X.
113. To reduce a compound fraction to its lowest terms.
We have seen that a simple fraction can often be reduced bj remov-
ing the factors common to the numerator and denominator. When the
common factors of a compound fraction can be detected by inspection,
we have but to remove them, and the fraction is reduced to its lowest
terms. Thus,
a^ + ah — ac
am -f- tt/t
can be reduced to its lowest terms by the removal of the common
factor rt.
But the common factor, or common factors, of a compound fraction
cannot always be detected by inspection, and some process becomes
necessary to discover them. This process is called finding the greatest
common divisor.
114. The greatest quantity that will divide two or more quantities,
is their greatest common divisor.
When the greatest common divisor of the numerator and denomina-
tor of a compound fractions is found, it can be reduced to its lowest
terms by dividing by this divisor.
115. The greatest common divisor, though most usually obtained in
order to reduce compound fractions, is also frequently found between
quantities not written in the fractional form.
116. We will then explain the method of finding the greatest com-
mon divisor between two polynomials, without regarding them as nume-
rator and denominator of a compound fraction.
117. The determination of the greatest common divisor of two poly-
nomials depends upon two principles. Ist. The common divisor of
two polynomials contains, as factors, all the common factors of the
two polynomials, and does not contain any other factors. For, let A
and B be the two polynomials, and D their greatest common divisor.
Now, it is plain that any quantity C, made up of a part of the common
factors of A and B, would divide both quantities, and would, therefore,
be a common divisor ; but C, multiplied by the remaining factors com-
mon to A and B, would still divide them. Hence, C would be a divi-
sor, but not the greatest divisor of A and B.
Again, any quantity M, made up of all, or a part, of the common fac-
ALGEBRAIC FBACTI0N6. CO
tors of A and B^ and containing also a factor, d, not common to the
two polynomials, will not divide them. For, if we proceed to the divi-
sion, the common factors of A, B, and M will strike out, and leave the
factor <f as a denominator of the reduced polynomials. Thus, ad will
not divide a* + <z^ ^d ac + am ; for, though it contains their common
factor, a, it also contains a factor cf, not common. The result of the
division will be — -r — and — -^ — . The greatest common divisor, D,
then, contains all the common factors of A and B, and contains no
other factors.
118. 2d. The greatest common divisor of two polynomials, A and 1^,
will enter into the successive remainders which arise from dividing A
by B, and B by the remainder, and the second remainder by the first,
and so on, until there is no remainder.
For, denote by A' and B' the quotients arising from dividing A and
B by the greatest common divisor, D ; then A = A'D, and B = B'D.
Divide A by B ; or, what is the same thing, divide A'D by B'D, and
call the quotient Q.
Thus, A'D I B'D
QB'D Q
(A' — QB') D =: 1st Remainder = MD,
by making A' — QB' = M.
We see that the first remainder contains the greatest common divisor;
and as it is also in the divisor, we have a right to seek it between these
polynomials.
Now, divide B'D by the remainder, and call the new quotient Q^,
and we get
B'D |MD
(B' — gM) D = 2d Remainder = XD,
representing B^ — (y^I by it.
We see that the greatest common divisor enters also into the second
remainder; and as it is also in the fii^st remainder, we have a right
to seek it between these two remainders. Divide MD by NDj the
remainder, if any, will still contain the greatest common divisor.
119. Let us apply these principles in finding the greatest common
divisor of the polynomials, A and B.
First arrange them with respect to a certain letter, and regard that
polynomial which contains the highest power of this letter as the divi-
70 ALGEBBAIC FRACTIONS.
(lend, and the other polynomial as the divisor. Next^ examine if A
(which we suppose the dividend) contains a factor common to all its
terms, but not common to all the terms of B. By the first principle,
this factor can constitute no part of the common divisor, and may be
suppressed. It is not absolutely necessary to suppress it, because it
will disappear in division, and not appear in the remainder, and, there-
fore, by the second principle, cannot make part of the greatest common
divisor. But if there is a factor common to all the terms of B, and not
common to those of A, it must be suppressed. For A would not be
divisible by B until it had been multiplied by the common factor of B,
and the multiplication would make this factor common to A and B;
and, hence, by the first principle, it must be common to the divisor.
The greatest common divisor would then contain a factor which did not
originally belong to A.
120. If there is a factor common to A and B which can be detected
by inspection, it may be divided out and set aside, as making part of
the greatest common divisor.
121. The next step, after setting aside, or suppressing factors seen
by inspection, is to divide A by B. If the coefficient of the first term
of A is not divisible by the coefficient of the first term of B, it may be
made divisible by multiplying all the terms of A by the coefficient of
the first term of B, or by any other quantity that will make the division
possible. This multiplication will not efifcct the result, because the
factor introduced into it will disappear in the division.
122. The remainder, after division of A by B, will contain the great-
est common divisor; and, as B also contains it by hypothesis, we have
a right to seek it between B and the remainder. The second remain-
der, if * any, contains D also, but its coefficient or multiplier is smaller
than in the first remainder; and so the coefficient of D, in each re-
mainder, is smaller than in the preceding, until it finally becomes unity.
When this final remainder is used as a divisor, it will go as many times
in the preceding divisor as there are units in the coefficient of D in
that divisor, and there will be no remainder.
123. When, therefore, we get a remainder zero, we conclude that
the last divisor is D X 1, or D itself.
124. If the given polynomials have no common divisor, we can dis-
cover the fact by the same tests that show when one polynomial is not
divisible, by another. We will find either that the letter, according to
which the arrangement has been made, has disappeared from some of
ALQEBRAIO FBACTI0N8. 71
the remainders, or it enters to a higher power in that remainder which
is used as a divisor than in the one used as a dividend.
The preceding principles and demonstrations for finding the greatesi
common divisor lead to the following
RULE.
125. Arrange the two polynomiah with reference to a certain letter^
as in division; ttse that one which contains the highest power of this
letter as the dividend,
2d. Next, set aside the factors^ if any, common to dividend and
divisor, as part of the greatest common divisor, and suppress those fac-
tors which are common to the one and not to tJie other,
Sd. Prepare the dividend, if necessary, for division, hy multiplying
by any quantity that will make its first term divisible by the first term
of the divisor ; divide the one polynomial by the other, and suppress
in the remainder any factor that m^y be common to all its term^ and
not common to those of the divisor,
4th. Use the remainder so reduced as a divisor, and the last divisor
as a dividend, and proceed a« before; and continue in this manner,
using each successive remainder as a divisor, and the last divisor as a
dividend, until there is no remainder, or until it is evident that the
polynomials have no common divisor,
--, , a'cb — aVy + a^bx — a^vx . .. ,
Eeduce — -. '-i ^^— to its lowest terms.
mco 4- ffico -+- ^nox + myx
a* is common to the numerator and not to the denominator, m is com-
mon to the denominator and not to the numerator. These factors must
then he suppressed, as constituting no part of the greatest common
divisor.
Then, cb — cy + bx — yx \ cb + cy -\- bx -}• yx
cb -{■ cy -}- bx -i- yx 1 Quotient.
— 2cy — 2yx = — 2y (c + x).
Suppress — 2y, a factor common to the remainder and not to the
divisor.
Then, cb + cy + bx + yx \ c -f x
cb -^ bx b •\- y 2d Quotient.
cy + yx
<y -h yx
+ 0.
72 ALQEBBAIO FBAGTIOKS.
Hence^ c -f- x is the greatest common divisor, and the reduced frac-
tion obtained by dividing both terms of the fraction by the Gt, C, "D, is
vn(b + t/)'
The factor suppressed in the first remainder might have been -f 2y,
instead of — 2i/. The greatest common divisor then would have been
— c — X, And, in general, the two polynomials have two greatest
common divisors, di£fering in the signs of all their terms. The reason
of this is obvious, since -rr-5 = — - , the common divisor to the two
terms of the fraction, may be, either + c7, or — d.
2. Find the greatest common divisor of a:* + 4a:? -j- Sx 4- 2, and
2x'-f 3x + !• Prepare for division by multiplying the first polyno-
mial by 4, the square of the coefficient of the first term of the second
polynomial.
Then, 4a:^ + 16a:^ + 20a; + 8
4a;^ + Gx" -f 2a;
10a;" + 18a; + 8
10x« + Ibx -f 5
2x" + 3a: + 1
2x + 5 Quotient.
3a; + 3 = 3 (a: 4- 1). Bemainder.
Suppress the common factor, 3, of the remainder, and continue the
division.
2a;» + 3a; + 1 | a; 4- 1
2a:» + 2a; 2a; + 1. 2d Quotient, a; + 1 == G.C.D.
x+ 1
a; + 1
126. If we had multiplied, to prepare for division, by the first in-
stead of the second power of the coefficient of the first term, we would
only have found one term in the quotient by the first division, and it
would have been necessary to multiply the remainder by the coefficient
of the first term to obtain a second term of the quotient. It is easier
to prepare the polynomial by multiplying by the second power of the
coefficient of the first term.
„ „ , a;* + 4ar» — 3x» — 10a;+ 8, . ,
3. Reduce rj— = ^ = to its lowest terms.
3ar — 4a; 4- 1
Multiply the numerator by 3' = 27, to prepare for division.
ALGEBRAIC TBACTI0N8. 78
27a^ + 108x' — 81x« — 270a; + 216 |8x'-^4ar-f 1
27x*— 36x'+ Ox* 9x* + 48x + 34 Quotient
144x» — 90a:' — 270aj + 216
144a:'— 192a:' -f 48a;
1020:"— 318.T; + 216
102a:'— 136x + 34
— 182x + 182 = — 182 (a; — 1).
Suppress — 182. i
Then, 3a;' — 4x + 1 | a; — 1
3a;' — 3x 3x — 1
— X + 1
— a; + 1
.
Hence, x — 1 is the divisor sought, and the reduced fraction
x'' -f Sx" 4- 2x — 8
3x — 1 •
127. In the above examples, the difference of the exponents of the
arranged letter being two, we multiply by the square of the coefficient
of the first term of the divisor to prepare the dividend for division. It
could have been prepared by multiplying by 3, but we would only have
gotten one term in the quotient, and there would have been two re-
mainders to prepare for division.
In general, when preparation for division is necessary, we save time
by multiplying the dividend by the coefficient of the first term of the
divisor, raised to a power one greater than the difference of exponents
of the arranged letter in the first terms of the two polynomials.
4. Find the greatest common divisor of
a* — a^x + aoi^ — a?, and a' — a'x + ax* — a^.
Ans. a — ^•
cr* + ca;* + a/' — x^ — x^ — a*^
5. Beduce
ca^ — ca^ -^ ci^ — a;* + x^ — x^'
Ans. GCBc — ^-
And reduced fraction, -^ = ^a-
of — x^ + 2r
7
74 ALGEBRAIC TEACTIONS.
6. Reduce ^~^'^^ , o^g~o^ to its lowest terms.
Am, (JCDx« — ^.
Reduced fraction, ^^-^ + ^/ + .ya:' + y'
c + 2
7. Reduce ^ ; J__ to its lowest tenns.
nar — Zanx + nx^
Ans, GCD a* — 2ax + a:«.
Reduced fraction, — — — .
n
8. Find the GCD of 3a;« — 3/ and 4x> + 4/.
9. Find the GCD of x* + a^ — aa^ + a«x — ax + a« and 7? — a^
— ax" -J- a*x + ax — a*.
u4»«. X* — ox + a*.
10. Find the GCD of x»« — ^'« and x' + x^ + xy« + f,
AnB. x* 4- ^.
11. Reduce — , ,/7 • to its lowest terms.
X" -f 4a.7;* + 5a*x + 2a'
Am. GCD X* + 2ax + al
x» -f 2ax + a*
Reduced fraction,
X + 2a
io -D J X* + x" + xV-' 4- xir* + 1 + X-*
12. Reduce -^ ^ ■ ^,,-» .-._x. -» ^ to its lowest terms.
X — x + xy — y +x * — a-*
^/w. GCD x-» + y' + X.
X "4" X
Reduced fraction, .
X — 1
13. Find the GCD of x> + x^ + 1 + ^ H-y + yx-» and x* + yx»
+ 0^+5^+/ + ^.
4w«. x* + y.
1 ^ -D J xV + 2x" + 3x + x?/ -f 1 . ,
14. Reduce — ^-- ^ — -L-I. to its lowest t^rma
xy + 2x + 1 + yx-» + j^« + 2y ™®*
^ns. GCD x-» + y + 2.
X^ J. 2J
Reduced fraction, -.
X -\- y
'""""° ""Tro»8.
Suppress factor a; "^ 1- W =
15. Beduce J-i-2C!jt£l+«-^/
+^^^T^'^"+^=^ to its West terms.
Educed fraction, ^+ ^-» • ^"'- <^D « + y».
128. Tie last ^''^''^'
S^test commnn J- • ''""•"P'es can be rB«d;i„ „ ^ , ,
the factor " , P^'^^o-^iak Suppre't ! '°''*'''' ^''^ «<«"'"0"
«nue i^I-*''^'"' «°<' divide the 0701"^^'": ""' '""^ «"^»g«<i !«*»«-
^^~2a, :,^
^ *~^ -^ as the divisAr *i,
— -2«4-2„- "'' *'^ '«'-^'"^- Will be
and * ° + ^, or aCa; — o-) ,
Md — a:* — 2a; . o ^ -; — («
^* + 8 -J == (a — 1) (j; 2>,
Snppress the fector a-T • ^
d: diw Of the ^;^;z^ ^--> — -
76 ALOEBBAIC FRACTIONS.
Use tlie first as the divisor; the three remainders will be
0, — 2a (x + 1) + X + 1 + a« (a; + 1), and 2a« (x + 1) — 2a (x + 1),
or 0, and (x + 1) (1 + a* — 2a), and (x + 1) (2a" — 2a).
Hence, x + 1 = GCD.
3d. Find the GCD of a« — x«, a" + 2ax -h x« and a* + 3a*x +
3ax*4" x*.
Am. a -f ar.
4th. Find the GCD of x^ — y% x» — y', x« — y'*, and x*' — y".
-4fi«. x* — y*.
5th. Find the GCD of x* — 1, x* — 1, x« — 2x + 1, and x» — 3x«
+ 3x — 1.
Arts, X — 1.
130. It Tfill be seen that the GCD of any number of polynomials, a8
well as for two, may have its sign changed. Thus, the last GCD may
be X — 1, or 1 — x, and so for the others.
LEAST COMMON MULTIPLE.
131. The least common multiple of two or more quantities, is the
least quantity that they will exactly divide. Thus, the least common
multiple of 2, 4, and 6, is 12; of a*, a, and ft, is a^h; of a* -f a&, h and
ft', is a'ft* + aft', &c. It is plain that the product of all the quantities
will be a common multiple of these quantities; that is, it will be exactly
divisible by them. Thus, 2x4x6 = 24 is a common multiple of 2,
4, and 6, but it is not their least common multiple. In like manner,
fi'ft is a common multiple of a', a, and ft, but not their least common
multiple. Multiply a common multiple by anything whatever, the pro-
duct will still be a common multiple. Hence, there may be an infinite
number of common multiples, but there can be but one least common
multiple.
132. No quantity tfill be divisible by another, unless it contains all
the factors of the second quantity. So, no quantity will be divisible
by two or more quantities, unless it is divisible by each, and by all the
factors of these quantities. To be the least common multiple of the
given quantities, it must contain no more factx)rs than they contain; and
these factors must not enter to higher powers than in the given quan-
tities. Thus, ahc is not the least common multiple of a and ft, because
ALOEBRAIC FRACTIONS. 77
it contains a factor, c, that they do not contain. Neither is a^h the
least common multiple, because the factor a enters to a higher power
than in the given quantities. The expressions ahc, and a'5; are multi-
ples, but not least common multiples.
133. If the given quantities contain a common factor, this must
enter into the least common multiple raised to the highest power to
which it is raised in the expressions to be divided, but it must not be
repeated.
134. It must enter to the highest power, else the multiple which we
formed would not be divisible by the expression containing the highest
power of the common factor, and it must must not be repeated, else the
multiple would not be the least common multiple.
135. From these principles we derive the following
RULE.
Decompose the given quantities into their prime factors, form a pro-
duct composed of all the factors not common, and of the highest powers
of the common factors, talcing care to let no factor enter more than
once. The product so formed will he tJie least common multiple
required.
Form the least common multiple of aa^, a^x, hx^y and a'.
Decomposing, we have a.a^, a^.x, h.x*, and a*.
Hence, least common multiple, c^.x*.h ss a^bji^.
It is pldn that the common factor, a, must enter to the highest
power (the third) to which it enters in one of the given quantities,
else the multiple would not be divisible by that quantity. It is also
plain that x must enter to the highest power, and that the factor b,
not common, must also form part of the least common multiple, other-
i^ise, the expression containing b would not be a divisor.
Form the least common multiple of 2, 8, 3, 9, a, a', and 55.
Decomposing into factors we have 2, 2', 3, 3', a, a", 5.6.
Hence^ 2^.3'.a'.&.5 ss 360a'& is the least common multiple required.
By inspecting the result, 2', 3', ^a\b, it is plain that it is the least
common multiple. It is divisible by 2, because it contains a &ioU>t 2';
7*
78 ALOSBRAIG FRACTIONS.
by 8, because it contains 2'; by 3, because it contains 3', &c. More-
over, it is the least product that will divide the given quantities, for it
is made up of the least factors that will fulfil the required conditions.
The number 2 need not be raised to the third power to divide 2, but
it must be to divide 8 ; so 3 need not be raised to the second power to
divide 3, but it must be to divide 9.
We see, too, that no factor has been taken more than once.
3d. Find the least common multiple of (a* -f crx^, 9a', 21, and lab.
An$, m (a + x') a^b = 63 (a + o^ a'^-
4th. Find the least common multiple of a-* — !^t ^ + y? 7x — 7y,
and a^ — i^x.
Ans. 7x (x* — y*).
*
5th. Find the least common multiple of 3aV, 4 (a; + 1)', 16a'?-»*,
and 40 (x + 1)&.
Ans, 3.2^5 (x + 1)^^^ = 240 (x + 1)VZ»».
6th. Find the least common multiple of 2a, 3a', 4Z>, 56', 6r, Tc*, 8c?%
9c/, and lOahcd,
Ans. 2'.3'.5.7a*6Vrf^ = 2520a*6Vrf\
Corollari/,
136. The least common multiple of two or more quantities can also
be found by dividing their product by the greatest common divisor.
For, in the division of the product of the quantities by their greatest
common divisor, the lowest powers of their common factors alone are
divided out, and the quotient is made up of the factors not common,
multiplied by those that are common, raised to the highest powers to
which they enter in any of the given quantities, which, as we have
seen, is the composition of the least common multiple.
137. Conversely, if we know the least common multiple of two or
more quantities, we can find their greatest common divisor by multi-
plying the quantities together, and dividing their product by the least
common multiple. For, let A, B, and C denote the quantities, D, their
greatest common divisor, and L their least common multiple. Then,
. ABC T -A.BC T. rrv 1 .. 1.
since —jT- = Ai> — y — = 1^- This relation between the greatest com-
mon divisor and the least common multiple, enables us to yerify our
lesults in finding either.
ALOEBBAIC FRACTIONS. 79
LEAST COMMON MULTIPLE OF FRACTIONS.
138. It is often important to find the least common multiple of frac-
tions, and as the role for finding it for entire quantities fails in this
case, it becomes necessary to demonstrate another rule.
Let us take the fractions
a' a' a*h
We are required to find the least quantity that they will exactly divide,
and ^ve entire quotients. Now, to divide by a fraction, is to divide
by the numerator, and multiply by the denominator. The quantity,
a'
then, that will be divisible by one of the fractions, as 7-.) T^^ist be divi-
* her
sible by a', or it will not be divisible after it has been multiplied by W.
And, as the same reasoning may be extended to the other fractions, the
quantity sought must evidently be divisible by each of the numerators;
and, in order to be the least quantity that will fulfil this condition, it
must be the least common multiple of the numerators. But, since the
denominators are to be multiplied by this least common multiple of the
numerators, it is plain that a!^b cannot be the quantity sought. For,
a*h, the least common multiple of the numerators, divided by any quan-
tity that will exactly divide the denominators, will be a smaller quantity
than a^b itself. For instance, — will be divisible by the given quan-
c
tities, and be a smaller quantity than a^h. It is plain, too, that a%
divided by the greatest quantity that will divide the denominators, will
still be smaller than — , and will be the smallest quantity that will be
exactly divisible by the given quantities. Hence, --^* is the least com-
mon multiple of the fractions
b?' w? "?•
RULE.
139. Find the least common multiple of the numerators, and divide
it hy the greatest common divisor o/tke denominators; the fraction so
formed will be the least common multiple of the given fractions.
80 ALGEBRAIC FRACTIONS.
2. Find the least common multiple of 7%, -fy^j and ^.
Ans. ^*.
3. Find the least common multiple of
S (a + a^ 2 (a» + as^) 4a
3 ' "9 ' "^"^ 27"
o
4. Find the least common multiple of
^ ^ , 4(a* — a«a:«) 2(a — ic) . 6 (a + a:)
2a, Ba% -^ — ^ ^, — ^^ ^, and g ^.
Am. 12a« (a« — a«).
5. Find the least common multiple of
7 (x+JT 49 ( x + y) 21a^ a; + y
3 ^ 9 ' '2r' T2"*
. 3 (49) (x + yy:c» .Q. ^ .,^
o
6. Find the least common multiple of
73a" 73a6 a; -f y , a
^««. 73a»6 (x + y).
The demonstration heing founded upon the hypothesis, that the frac-
tions are reduced to their lowest terms, the rule is, of course, only
applicable to such fractions.
GREATEST COMMON DIVISOR OF FRACTIONS.
140. The greatest common divisor of two or more fractions is the
greatest quantity that will exactly divide them, giving entire quotients.
This quantity must be a fraction ; for nothing but a fraction will-
divide a fraction reduced to its lowest terms, and give an entire quo-
tient. The gre&test common divisor of several fractions will then be a
fraction itself; and since, when we divide a fraction by another frac-
tion, we divide the numerator of the fraction assumed as the dividend
by the numerator of the fraction taken as the divisor, and divide the
denominator of the divisor by the denominator of the dividend, it fol-
lows, that the divisor sought must have a numerator that will divide
each of the numerators of the given fractions, and a denominator that
AIiQEBBAIO PBACTIONS. 81
will be divisible by eacb of the given denominators. But, since the
value of a fraction increases witb the increase of its numerator and the
decrease of its denominator, it is plain that a fraction, whose numerator
would divide all the given numerators, and whose denominator would
be divisible by all the given denominators^ would not be the greatest
fraction that will divide the given fractions, unless it has the greatest
numerator and the least denominator that will fulfil the required
conditions. Or, in other words, unless its numerator is the greatest
common divisor of the given numerators, and its denominator the leaat
common multiple of the given denominators.
RULE.
141. Find ihegreateit common divisor of the numerators, and di-
vide it hy the least common multiple of the denominators; the fraction
90 formed vsiU he the least common multiple of the given fractions.
EXAMPLES.
1. Find the GCD of ^, ^', ^, and 6a«. Ans. ^.
« -n. -I .^ /^i^-rx i. 7x' 5x' (a-\'X)x . X* . x
2. Find the GCD of ;n7> tt-j a > and -^. Ans. .r^.
12 3 4 2 12
2a S 31? X 1
3. Find the GCD of -5-, -p, =7, and 53. Ans. ^.
4. Find the GCD of jr-, -^, — , and — . Ans. ^ — .
2a 6 X X Dox
5. Find the GCD of -=-, ^, ^ , and -^. Ans. ^.
6. Find the GCD of 4", ^, , ^. ,, and -3.
*2x x' a'x + or or
Ans,
12x' (a« + x)'
Remarks.
142. The greatest common divisor of entire quantities may be unity,
and the quantities are then said to be prime with respect to each other.
But it is evident that fractions can never be prime with respect to each
82 EQUATIONS OF THE FIR8T DEQBEE.
other ; for, though their numerators may be prime^ s& in the 3d, 4thy
and 6th examples, the least common multiple of their denominators can
always he formed.
143. We may also remark that, as the least common multiple of
entire quantities can always be formed ; so, likewise, all fractions what-
ever have a least common multiple.
144. Knowing the least common multiple of any number of frac-
tions, we can find their greatest common divisor by multiplying the
fractions together, and dividing their product by their least common
multiple.
145. Conversely, knowing their greatest common divisor, we can
find their least common multiple by multiplying the fractions together,
and dividing the product by the greatest common divisor.
EQUATIONS OF THE FIRST DEGREE.
146. An Equation is an expression containing two equal quantities,
with the sign of equality between them. Thus, ar = a — 6 is an equa-
tion, and expresses, that the quantity represented by x is equal to the
difference of the quantities represented by a and b.
147. The part on the left of the sign of equality is called the^ir*;
member of the equation, and that on the right the second member.
148. Problems can be stated or expressed, and their solutions ob-
tained by mciins of equations ; that is, we can express, in algebraic lan-
guage, the relation between an unknown quantity to be found, and one
or more known quantities, and, by certain operations, can find the value
of the unknown quantity.
149. The expressing the relation is called the statement of the pro-
blem; and the operation performed after the statement, to find the
value of the unknown quantity, is called the solution of the problem.
Thus, let it bo required to find a quantity, which, being added once to
itself, will give a sum equal to b. Let x be the unknown quantity;
then, by the conditions, aj + a; = ft, or, 2x = b. Then, if twice x is
equal to 6, x itself must be equal to half of b, and a? = -- is the final
result. In this, making the equation, x + x =^ b is the statoment.
EQUATIONS OF THE FIRST DEGREE. 88
and the sabsequent operation is the solution. If the value found for
the unknown quantitity ( -r ) ^ substituted in the equation x + xssh^
we will have —-+-—== fc, a true equation. When the value found
for the unknown quantity substituted in the equation of the probleifi
makes the two members equal to each other, the equation is said to be
tatUJUd, and we conclude that the solution is true.
150. The unknown quantity, the thing to be found, is usually repre-
sented by one of the final letters of the alphabet, x, y^ and z. Known
quantities are generally represented by the first letters of the alphabet;
a, hj Cj and d,
151. An equation with one unknown quantity is of the first degree,
when the highest exponent of the unknown quantity in any term is
unity; of the second, when the highest exponent of the unknown
quantity in any term is two, &c.
a: + a 3= 5 is an equation of the first degree.
x* + a: =s a is an equation of the second degree.
3^ + 3:* + ^ = ^ 18 an equation of the third degree.
X* + j?a;' + »*x' + wo: = a is an equation of the fourth degree.
152. Equations which contain the unknovm quantity from the high-
est to the first power inclusive, are called complete equations. The
above are all complete equations with one unknown quantity.
153. Equations in which some of the powers of the unknown quan-
tity are missing, are called incomplete equations, a;" s* a is an incom-
plete equation of the second degree, a^ + x* s a is an incomplete
equation of the third degree.
154. A numerical equation is one in which all the known quantities
are represented by numbers. Thus, x* + 2x = 4 is a numerical
equation.
155. A literal equation is one in which all the quantities, known
and unknown, are represented by letters. Thus, x' -f ax = ft is a
literal equation.
156. An equation in which the known quantities are partly repre-
sented by letters, and partly by numbers, is a mixed equation. Thus,
x'-haaf=s6-f2isa mixed equation.
84 EQUATIONS OF THE FIRST DEGREE.
157. Ad identical equation is one in which the two members differ
only in form, if they differ at all. Thus, 2x = ~, — 5 — ^x •{- -5,
and X =s X, are identical equations.
158. An indeterminate equation is one in which the value of the
unknown quantity is indeterminate.
159. A single equation with two unknown quantities is necessarily
indeterminate, for we must assume the value of one before we can
determine that of the other. Thus, x + 3/ = 10 is an indeterminate
equation, because, by assuming y = 1, 5, 4, &c., we find x =1 9, 6, 6,
&c. And, by attributing an infinite number of arbitrary values to y,
there will be an infinite number of values for x.
160. All identical equations are necessarily indeterminate. Thus,
the equation x = x will be true, or satisfied, when x is 1, 10, lOOO, or
anything whatever.
161. The 0W71 sign of a quantity is the sign with which the quantity
is affected previous to any operation being performed upon it. The
essential sign is the sign with which it is affected after the operation.
Thus, multiply + a by — 1 ; the own sign of a is plus, and its essen-
tial sign minus. But, multiply + a by + 1, and the own and essen-
tial signs are the same. Subtract + h from a, then a — (-f ft) sbs= a
— b. Here, b appears in the second member with the essential sign.
•162. Since a positive quantity cannot be equal to a negative, and,
conversely, it is evident that the essential sign of the two members of
an equation must be the same.
163. Since quantities can only be equal to quantities of the same
kind, it is plain that the two members of an equation must be composed
of quantities of the same kind. Thus, if one member represent time,
the other must represent time also.
These two principles are important, and ought to be remembered.
164. Several axioms, or self-evident propositions, are assumed as the
basis of the principles by which equations are solved.
1. If equals be added to or subtracted from equals, the results will
be equal.
2. If equals be multiplied or divided by equals, the results will be
equal.
EQUATIONS OP THB PIEST DBOREE. 85
SOLUTION OF EQUATIONS OF THE FIRST DEGREE.
165. The transformation of an equation consists in changing its form,
without destroying the equality of the two members.
There are three transformations of equations.
First Tratis/ormation.
1G6. The first transformation consists in clearing an equation of its
fractions.
By the second axiom, we have a right to multiply both members of
an equation by the same quantity ; and it is evident that if we multi-
ply the two members by the least common multiple of the denominators,"
the resulting equation will be free of fractions. For, since each deno-
minator will divide the least common multiple, it is plain that the pro-
duct of each numerator by the least common multiple will be divisible
by each denominator. Thus, take the equation — -f - -f 2 = 4, mul-
tiply each member by 6, the least common multiple of the denomina-
tors, and the resulting equation is 3x -f x + 12 = 24.
It is plain, that if we multiply the two members by 12, the product
of the denominators, the resulting equation will also be free from frac-
tions. For each denominator will divide the product of 12 by its
numerator, inasmuch as it will divide 12 itself. Multiplying by 12,
we get 6a; + 2x + 24 = 48. Now, divide both members by 2, which
we have a right to do by the second axiom, and we get 3x -t- a: = 12,
the same equation as before.
OC OC 3C
Again, take the equation rt- + -ft+x"^^~^'
By the first method (the least common multiple being 12), we get
Gx + 2x + 3x + 24 = 60. By the second method, we get 24x + Sx
-f 12x -f 96 = 240. Divide both members by 4, and the results are
the same. It will be seen that the second method is the same as mul-
tiplying each numerator into aU the denominators except its own, and
each entire quantity by the product of all the denominators.
RULE.
Multiply both members hy the least common multiple^ or by the pro-
duct of the denominators,
8
86 EQUATIONB OF THE FIRST DEOBES.
When the least common multiple can be found without difficultyi
the first method is preferable.
Second Transformation,
167. This consists in transposing known terms to the second mem-
ber^ and unknown terms to the first member.
Take the equation 2x — 2 = a: + a. (1)
By the first axiom, we have a right to add -f 2 to both members of
the equation. Adding, we have 2x — 2 + 2B=sa; + a-f2; or, since
the — 2 and + 2 destroy each other in tlie first member, the equation
becomes 2x = x -\- a + 2. By comparing this with the equation
marked (1), we see that 2 has passed into the second member, by
changing its sign.
Kesume the equation 2x = x + a + 2, and add minus x to both
members. We got 2x — x s^ x — x + a+2; or, since + x and
— x destroy each other in the second member, 2x — x ^ a + 2.
Hence, x has passed into the first member by changing its sign.
We have demonstrated that quantities can be transposed from one
member to another, provided we change their signs, since transposition
is nothing more than adding to both members the quantities to be
transposed, with their signs changed. The demonstration can be as
readily made by subtracting the quantities with their appropriate signs.
Hence, the se6ond transformation is equivalent to adding to both mem-
bers the quantities to be transposed, with their signs changed, or sub-
tracting these quantities with their own signs from both members.
RULE.
Change the zign of each term trans/erred from one nieviiber to
another.
Third Trantformation.
168. The third transformation consists in clearing the unknown
quantity of its coefficient or coefficients.
Resume the equation, Cx + 2a: + 3x H 24 = 60 ; then, by the second
transformation, we have 6j; -f 2a: + 3x == 60 — 24 = 36. This indi-
cates that the sum of 6x, 2x, and 3x is equal to 36 ; hence, obviously,
llx =s 36. Now, by the second axiom, we have a right to divide both
EQUATIONS OF THB FIRST DEQBBE. 87
members by the same quantity. Divido by 11, and we have x ss |fi.
The unknown quantity now stands unconnected with known termSi and
we have its true value.
Take ax + hx — ex =» m.
By the rules for factorings we have (a + ft — c)a; = m; and by
the second axiom, x =s — .
a + — c
These illustrations show that the third transformation is used when
the equation has been cleared of its fractions, if it contained any, and
when all the known quantities in the first member have been removed
to the second, and when all the terms involving the unknown quan-
tity have been placed in the first member, if not already there.
RULE.
Collect into a single algebraic sum all the coefficients of the unJcnovm
quantitj/j and divide the two members bi/ this sum; if there is but one
coefficient, divide both members bi/ it,
EXAMPLES.
1. Clear the unknown quantity of its coefficient in the equation,
2x = 6 -f c
Ans. X a= — - — .
2. Clear the unknown quantity of its coefficient in the equation,
6x + 3x — ex = a.
Ans. X :s
b + B — c
8. Clear the unknown quantity of its coefficient in the equation,
bx + ax — 2x=^n.
Ans. X =
3+a
4. Clear the unknown quantity of its coefficient in the equation,
7x + 3a; 4- 2x = 12.
Ans. X sssl,
5. Clear the unknown quantity of its coefficient in the equation,
2x — 7x = a.
. a — a
Ans. X = — =: ^ — ^ .
— 5
88 EQUATIONS OF THE FIRST DEGREE.
169. It will be seen tKat the object of the three transformations is
to free the unknown quantity of its connection with known terms, and
make it stand alone in the first member of the equation. When so
situated, its true value is known, being expressed by the second mem-
ber of the equation.
170. The various steps in the solution of an equation, can be best
illustrated by taking an example which will require all three trans-
formations, f
Take |. + ?f +a=:6_2x.
First step. To clear the equation of its fractions.
EesulL 5:c H- 3a; + 10a = 105 — 20x.
Second step. To get the known terms to second member, and un-
known to first member.
JResulL 6x 4- 3x + 20a; = 10b — 10a.
Third step. To clear the unknown quantity of its coefficient.
zH 14
171. Frequently, only two transformations are necessary in the solu-
tion of an equation ; sometimes only one, but always at least one.
Take ax + h s= c.
c h
By second transformation, ax = c — h; and by third, x = .
Take 2x — 3x = w ; then, by third transformation, — x = m.
In this the unknown quantity appears with the negative sign ; but
we were required to find the value of + a; and not of — x. Multiply
both members by minus unity, which we have a right to do by the
second axiom, and — x becomes positive. Hence, x == — m is the
required value.
172. In general, when the sign of the unknown quantity is negative
in the final result, and the conditions of the problem require it to be
affected with the positive sign, both members must be multiplied by
minus unity.
173. We observe, also, that the signs of all the terms of an equation
BQUATIONS OF THS FIRST DXOREB. 89
can be changed without altering the equality of the two members^ since
we have a right to multiply these members by minus unity.
174. From the foregoing principles we derive, for the solution of
equations of the first degree, the following
RULE.
I. If the equation contain fraction^f clear it of them hi/ multiplying
both members by the product oftJte denominators^ or their least common
multiple,
II. Bring all the terms involving the unknovm quantity to the first
member (ifn(^t already there"), and all the known terms to the second.
III. Collect into a single algebraic sum all the coejfftdenfs of the un^
known quantity, and divide both members by this sum.
TV. If the unknown quantity appear in the final result with a nega-
tive sign, multiply both members by minus unity.
EXAMPLES.
1. Solve the equation a; + 3x — }x = 14. Ans. x s 4.
2. Solve the equation Ax + -^ — ttt + -?- =60.
Ans. X s= 10.
X XX
3. Solve the equation -f- + ^^ + tT+ -q- — ^0 = — 6.
Ans. X = 15.
4. Solve the equation 2x + Sx + 4x — 7 = — |.
Ans. X rss i.
X X X X X 24^ 176
5. Solve the equation y + -^^ + — + -^- + _ ^= _-.
Ans. X = 12.
6. Solve the equation 2x + Sx + ix — 7 = — 4.
Ans. 05 «= J.
XX c
7. Solve the equation 1 \- Sx ^^ 1- Sc + 1.
a a
Ans. X s: c.
. 8*
90 EQUATIONS OP THE FIRST DEGREE.
X
1?
8. Solve the equation ixx -\ — , + 4a; — 6 = n' + 4 (w* — 1).
Anz. X = n\
X
9. Solve the equation — -- + ix + h — a = 5& -f 3a + 1.
a + 6
Ans, X ^ a + b.
abed
10. Solve the equation hxH 5 = ^^^^•
A (abcdy
bed + acd + abd — abc
11. Solve the equation ~ j- -i 1 = 4-
2 4 j3 m a
(j3^ +pm + pq + 4y + 4m + 4a) p + a __ (p + m)
4p "*" m a '
Ans, a: = p 4- m + a.
12. Solve the equation z -\ 1- a^ + az = m,
a
. a (m — ab)
«' + a + 1
13. Solve the equation t/ — ^ + ^4- a + by = a+ ~ c.
1 o 3
— 105c
111 + 366"
Ans. y =:
14. Solve the equation -/ + i^ — ^+a + 3 — 6i>==10.
o a
A c (7 + 65 — a')
175. A few examples follow of equations in which the unknown
quantity is affected with negative exponents, and in which the degree
of the equation is apparently higher than the first.
1 a 1 av X- 2x-' + a; — 1 2a; + 4
1. Solve the equation — ss — -L .
3 6x
Ans, a; s=s 2.
3
2. Solve the equation ^ — f- ^ = 2.
^^ Ans, a; = J.
8. Solve the equation ~ -f -— = ?—- •
^5 5 ^w«. a; = 5.
EQUATIONS OF THE FIB6T DEGREE. 91
A Qj .1. *• a^-hTa^— 3x 12x« + 14a;'
4. Solve the equation jr = 71 x-
Am, a; sa 6.
/jj-i X 2x'~^ 1
5. Solve the equation r-r + ■;; =— = ;; — •• a a
^ 14 7 7 7x~* Ans. as = 4.
6. Solve the equation x"^ + — H — = =—5 .
QiX X OX
Ans. X 39 -p.
6
7. Solve the equation l^-l + J-f^'^ + J^ 80a;-*.
Ans. X =s 40.
8. Solve the equation ax~* + — = ^ -v
a X or
^n«. X =s — = ;— .
1 — ab
3 4 8
9. Solve the equation 2a;-* + — | + -^ = x"* + -j^.
^n«. a; &3 1.
10. Solve the equation 2x"* = —5 h --r-.
o ar
^ Ans, a; aa a -f »i.
176. Solufions 0/ problems producing equations of the first degree
with an unknown quantity.
It has heen remarked, that the solution of a problem consists of two
distinct parts, the statement and the solution.
The examples just given, illustrate the manner of solving the equa-
tion after the statement has been made.
No general rules can be given for making the statement, which de-
pends mainly upon the ingenuity of the student.
Sometimes there is but a single thing to be determined, and in that
case, by representing it by one of the final letters of the alphabet, and
following the conditions of the problem, we get the equation or state-
ment required.
Often, however, two or more things are to be determined, then the
undetermined quantity, upon which the other quantities depend, is re-
presented by one of the final letters of the alphabet.
As an example of the first class of problems, let it be required to
find a quantity, which, added to half itself, will give a sum equal
to 6.
92 EQUATIONS OF THE FIRST DEGREE.
Let X represent the quantity, then half of the quantity will be repre-
sented by Jar, and by the conditions we have x + }x = 6. Solving
the equation, we get x = 4.
As an example of the second class of problems, take the following :
A man invests half of his money in State stocks, and a third of the
remainder in a Savings' Institution, and has four hundred dollars left.
Eequired the entire amount of his capital, and the amount of each of
his two investments.
Since his investments depend upon his capital, let x = capital.
X ~"~ \x
Then }x == amount invested in State stock, and — -^ — = ix= amount
o
placed in Savings' Institution. The remainder, after these investments,
is plainly x — (Jx + ix); but this remainder is, by the conditions of
the problem, equal to four hundred dollars. Hence, x — (§x + ix)
= 400. Solving, we get x = 1200.
Then, }x ss 600, Investment in State stock.
ix = 200, " in Savings' Institution.
400, The amount uninvested.
1200, Original capital.
The solution has been verified by adding the two investments to
what waa loft, and getting a sum equal to the original capital. The
solutions of all problems ought to be verified in a similar manner.
We have seen that an equation is said to be satisfied when the value
found for the unknown quantity substituted in the given equation,
makes the two members equal to each other
But the value of the unknown quantity in the solution of a problem
must not only satisfy the equation of the problem (the statement), but
must also fulfil the required conditions. In consequence of the use of
negative quantities in Algebra, the equation of the problem (the state-
ment), is often satisfied, when the required conditions are not fulfilled
in a strict, arithmetical sense.
As an illustration, let it be required to find a quantity, which, when
added to 4, will give a sum equal to 2. Let x be the quantity, then
X + 4 «= 2. Hence, x = — 2. Now, — 2 will satisfy the equation
of the problem, but will not fulfil the conditions in an arithmetical
sense; for we were required to find a quantity, which, when add^l to
4, would give a sum equal to 2, and the quantity found is really to be
taken from 4.
EQUATIONS OF THE FIRST DEUREE. OS
It will be seen hereafter, that there are several cases in which the
equation may be satisfied, and the conditions not fulfilled.
EXAMPLES.
A farmer has two kinds of oats ; the one kind worth 30 cents per
bushel ; tlie other 20 cents per bushel. He mixes 50 bushels of the
superior article with a certain amount of the inferior, and sells the mix-
ture at 22 cents per- bushel. IIow much of the second quality did
he put with the first in order to make the rate of sale the same ?
Let X represent the amount of the inferior article ; then, by the con-
dition, 60 . 30 -h 20a; = (50 + x) 22. Hence, x = 200. In this
example, the verification of the equation and the fulfilment of the con-
ditions are the same.
2. A farmer has 50 bushels of oats, worth 30 cents per bushel, and
200 bushels, worth 20 cents per bushel. He mixes the two lots to-
gether. At what price ought he to sell the mixture ?
Let X represent the required price of the mixture per bushel; then,
50 . 30 -f- 20 . 200 = 250x. Hence, x = 22.
3. What number must be subtracted from the numerator and deno-
minator of the fraction }, to make the new fraction the reciprocal of
the old ? An%, x = 7.
4. What number must be subtracted from the numerator and denomi-
nator of any fraction — to make the new fraction, the reciprocal of the
old. Am. a; ss m + n.
The solution is general, and is true for any fraction whatever.
5. What number, added to the numerator and denominator of the
fraction I, will make the new fraction 4 times as great as the old ?
An%, X = — 3*.
Wliat does this value satisfy ?
6. What number, added to the numerator and denominator of any
fraction ~, will make the new fraction 4 times as great as the old ?
. 3win
Ant, X s
n — 4m'
7. What number, added to any fraction — , will give a sum equal
to the numerator. . m (n — 1)
Am, X = — ^^ .
n
94 EQUATIONS OF THE FIRST DEOUSE.
8. What number, added to any fraction — , will give a sum equal to
the denominator ? . n* — m
n
Tn what case will this result fail to satisfy the conditions of the
problem?
9. What number, multiplied by any fraction — , will give a product
equal to the sum of the numerator and denominator?
(711 + n) n
m
How may the last four results be made applicable to any fraction
whatever ?
10. In a square floor of a college building there is a certain number
of brick ; if one more brick is added to each side of the floor, and the
square form be preserved, there will be 61 more brick than at first.
How many brick does the floor contain ? Arts, 3? = 900.
11. Two pipes lead into a reservoir capable of containing 1(>0 hogs-
heads of water. The first pipe can fill it in 16 hours, and the two to-
gether in 12 hours. In what time can the second alone fill it?
Ans, X = 48 hours.
12. Two travellers travel at the same Tate, the one starting before
the other 5 at 12 o'clock the first has travelled 4 times as far as the
second, but when they had each gone 15 miles further, the entire dis-
tance passed over by the first was only double that of the second.
Required these distances. Ans, 45 and 22 i miles.
12. The sum of two numbers is a, and their difference h. Required
the two numbers.
Arts. The greater, o- + o-j ^^® smaller -^ — — .
We see from this result that, knowing the sum and difference of two
quantities, we get the greater by adding the half sum to the half diffe-
rence, and the less, by subtracting the half difference from the half
sum.
This formula is of extensive application, and ought to be remem-
bered.
13. A Californian gold digger wishes to sell a vessel full of gold
mixed with sand. The vessel, when filled with gold, will weigh ten
pounds, and when filled with sand, one pound; the mixture weighs seven
pounds. How much gold is in it ? An%, 6 j\ lbs.
QEOMETRICAIi PROPORTION. 95
14. Tbe railing aronnd the altar of the Cathedral in the City of
Mexico is a composition of gold and silver. Assuming that 279 pounds
of the composition loses 20 pounds when immersed in water, and that
19 i pounds of gold lose one pound in water, and 10 i pounds of silver
lose one pound in water. Required the proportion of gold and silver
in the alloy. Ans. Gold : Silver : : 156 : 123.
15. Two men purchase together a barrel of flour, weighing 196
pounds, for 5 dollars ; the first pays half a dollar more than the second.
How ought they to divide the flour f
Ans. The first ought to have 107 J lbs, the second 88^.
GEOMETRICAL PROPORTION.
177. Eatio is the quotient arising from dividing one quantity by
another of the same kind. Thus, if M and N represent quantities
of the same kind, then ^ expresses the ratio of M to N.
Four quantities, M, N, P, and Q, are said to be proportional when
the ratio of the first to the second is the same as that of the third to
the fourth.
Thus, 64, 8, 16, and 2 constitute a proportion, because the first,
divided by the second, is equal to the third divided by the fourth.
The proportionality of four quantities, M, N, P, and Q, is expressed
thus, M : N : : P : Q, and is read, M is to N as P is to Q.
The first and last terms of a proportion are called the extremes ; the
second and third the means. Of four proportional quantities, the first
and third are called antecedents ; the other two, consequents. Of three
quantities, M, N, and P ; when M : N : : N : P, then N is said to be
a mean proportional between M and P ; and N is, at the same time,
antecedent and consequent.
Two quantities, M and N, are said to be reciprocally proportional,
when the one increases as fast as the other diminishes. One of the
quantities must be equal to a fraction with a constant numerator, and
with the other quantity for its denominator.
4
Thus, M = -i^ expresses that M and N are reciprocally propor-
tional.
96 QEOMETEICAL PROPORTION.
Equi-multiples of two quantities arc the products which arise from
multiplying them by the same quantity. Thus^ aM and aN are equi-
multiples of M and N, the common factor being a.
Theorem I.
If four quantities are in proportion, the product of the extremes will
equal the product of the means.
M P
For, when M : N : : P : Q, we know that i^ ==* 77 ; wid by clearing
of fractions, MQ = PN.
This theorem furnishes an important test of the proportionality of
four quantities. Whenever the product of the extremes is equal to
that of the means, the proportion is true ; and when that is not the
case, it is a false proportion.
Theorem II.
178. When the product of two quantities is equal to the product of
two other quantities, two of them may bo taken as the extremes, and
two as the means of a proportion.
For, suppose MQ = NP, divide both members by QN, the first mem-
ber will become rr-, and the second 77, and we have ^ = 77. From
N Q N Q
which we get the proportion M : N : : P : Q.
Let 2 . 4 »= 8 . 1. Then 2 : 8 : : 1 : 4.
Theorem III.
179. The squaro of a mean proportional is equal to the product of
the other two terms of the proportion.
For, from the definition of a mean proportional, we have M : N : :
N : P J hence, by Theorem I., N« = MP.
Let 2 : 4 : : 4 : 8. Then 4^ = 2 . 8 = 16.
Theorem IV.
180. When four quantities aro in proportion, they will also be in
proportion by alternation ; that is, when antecedent is compared with
antecedent, and consequent with consequent.
For, let M : N : : P : Q; then, by Theorem I., MQ = NP, and
OEOMETRIGAL PROPORTION. 97
M N
diyiding this equation^ member by member, by PQ, we get ^n "^ ?: »
1 (J
from whicb M : P : : N : Q.
Let 4 : 8 : : 12 : 24. Then, also, 4 : 12 : : 8 : 24.
Theorem V.
181. When four quantities are in proportion, they will be in propor-
tion when taken inversely; that is, when the consequents take the
place of antecedents, and the antecedents the place of consequents.
Let M : N : : P : Q; then, also, we have MQ = NP, or NP«MQ;
N Q
divide both members by MP, and we get ^ = p-> fro^i which N :
M : : Q : P.
Theorem VI.
182. If four quantities are in proportion, they will be in proportion
by compontion ; that is when the sum of antecedent and consequent is
compared with either antecedent or consequent.
M P
Let M : N : : P : Q : then, also, qrr "« --. Add 1 to both members,
N Q
ji A ^ ' . .,M + NP+Q
and reduce to a common denominator, and we have — r= — ss — ^ — ,
' N Q '
from which we get M + N : N : : P + Q : Q.
Let 4 : 12 : : 8 : 24. Then 4 + 12 : 12 : 8 + 24 : 24.
In like manner it may be shown, that when M : N : : P : Q, that
we will also have M4-N:M::P + Q:P.
Theorem VII.
183. When four quantities are in proportion, they will also be in
proportion by division ; that is, when the difference of antecedent and
consequent is compared with either antecedent or consequent.
M P
Let M : N : : P : Q : then, also, we have rr = tt* Subtract 1 from
N Q
both members, and reduce to a common denominator, we will have
M — N P— Q
— =j — = — TT — , from which we get M — N : N : : P — Q : Q.
Let 12 : 4 : : 24 : 8. Then 12 — 4 : 4 : : 24 — 8 : 8.
It may be shown in like manner, that when four quantities are in
7 o
98 QEOMET&IGAL PROPORTION.
proportion, M : N : : P : Q, that we will also have M — N : M : :
P— Q:P.
Let 12 : 4 : : 24 : 8. Then, also, 12 — 4 : 12 : : 24 — 8 : 24.
Theorem VIII.
184. Eqni-multiples of any two quantities are proportional to the
quantities themselves.
For, take the identical proportion, M : N : : M : N ; then MN =
NM. Multiply both members by m, and there results mM . N »=
mN . M ; and, since mM and mN may be regarded as single terms, we
have from Theorem 2d, M : N : : mM : mN.
Let 1 and 2 be multiplied by the same number, 3 ; then 1.2::
1.8:2.3; or 1 . 2 : : 3 : 6.
Theorem IX.
185. If equi-multiplcs of the antecedents of four proportional quan-
tities, and also equi-multiplcs of the consequents be taken, the four
resulting quantities will be proportional.
For, let M : N : : P : Q; then MQ = NP. Multiplying both mem-
bers by mw, and there results mM . wQ = «N . mP. Hence, mM :
nN : : mP : nQ.
Let 1 : 2 : : 4 : 8, and take m = 4, and n = S,
Then 4 . 1 : 3 . 2 : : 4 . 4 : 3 . 8; or 4 : 6 : : 16 : 24.
It is plain that the above theorem is equally true when m ss n. So,
that all the terms of a proportion may be multiplied by the same quan-
tity without destroying the proportionality of the terms. An infinite
number of proportions may then be formed from a single proportion.
Theorem X.
186. If there be two sets of four proportional quantities, having two
terms, the same in both, the remaining terms will constitute a pro>
portion.
Let M : N : : P : Q.
M : N : : R : S.
Then, MQ = NP, or NP = MQ.
MS = NR and NR = MS.
OSOMETRIGAL FBOPOBTION. 99
P Q
DiyidiDg these equations member by memDer^ we get ^ = 5-; from
which, P : R : : Q : S, or P : Q : : R : S. (Art. 180).
And the same property can evidentlj be shown when any other two
terms are equal.
Let 1 : 2 : : 4 : 8.
1 : 2 : : 6 : 12. Then 4 : 6 : : 8 : 12, or 4 : 8 : : G : 12.
Let 1 : 2 : : 4^ 8.
1 : 3 : : 4 : 12. Then 2 : 3 : : 8 : 12, or 2 : 8 : : 3 : 12.
It will be seen that the corresponding terms of the proportion must
be taken in connection with each other.
Theorem XI.
187. Of four proportional quantities, if the two antecedents be aug-
mented or diminished by quantities which are proportional to the two
consequents, the resulting quantities will be proportional to the con-
sequents.
For, let M : N : : P : Q.
N : Q : : R : S.
Then, MQ = NP. (A).
NS =QR.
Or, QR = NS. (B).
Adding and subtracting (B) from (A), there results Q (M =h R) =
N (P ± S). Hence, M±R:P±S:N:Q.
Let 1 : 2 : : 4 : 8.
And 3 : 6 : : 4 : 8. Then I=h3:2db6::4:8.
It can be shown in like manner, that if the two consequents be aug-
mented or diminished by quantities, which are proportional to the
antecedents, the resulting quantities will be proportional to the ante-
cedents.
The reciprocal of a quantity is unity divided by the quantity ; thus,
the reciprocal of A is -j .
Theorem XII.
188. Two quantities are inversely proportional to their reciprocals.
For, take the identical proportion A : B : : A : B ; then AB =s BA.
100 QEOMETRIGAL PROPORTION.
Divide both members by AB ; tben
Hence, A : B : : =r- : -7-. Thus, 2 : 3 : : i : }.
B A
Theorem XIII.
189. If there are any number of proportions having the same ratio,
any one antecedent will be to its consequent as the sum of all the ante-
oedents to the sum of all the consequents.
For, let M : N : : P : Q. '^
M:N::A:B. K
M : N : : C : D. [
M : N : : E : F. ,
Then, MQ « NP.
MB = NA.
MD = NO.
MF = NB.
Adding these equations member by member, we get M (Q + B +
D + F) = N (P + A + C + E).
Hence, M:N::P + A + C+E:Q + B + D + F.
Let 1:2:
1:2:
1:2:
4:8.
6:12.
8:16.
Then, 1 : 2 : : 4 + 6 + 8 : 8 + 12 + 16, or 1 . 2 : : 18 : 36.
It is not necessary that the first antecedent and consequent of each
proportion should be M and N; all that is required is, that they have
the same ratio; for, then, we can substitute M and N for them, and
the above demonstration becomes applicable. To show our authority
for this substitution, take the two proportions,
M : N : : P : Q, and R : S : T : U.
' R T
From the second, we get RU = ST, or ■=-====• ; and, since, by hy-
R M M T
pothesis, -^ =s =^, there results t^ = ff- Hence, M : N : : T : U.
GXOMXTBIOAL PBOPOBTION. 101
When we have any number of proportions, like those marked A,
having a common ratio, we can obTiously form a continued proportion
from them.
Thus, M : N : : P : Q : : A : B : : C : D : : E : F.
So, also, 1 : 2 : : 4 : 8 : : 6 : 12 : : 8 : 16.
The character : : is written before each new antecedent, and refers
it and its consequent back to the first antecedent and consequent.
It is obvious that any antecedent may be repeated any number of
times, provided, that its consequent is repeated the same number of
times ; and, also, that any consequent may be multiplied or divided by
anything whatever, provided, that the same operation be performed on
its consequent.
Thus, M:N::P + A4-C + E:Q + B-fD + F,
may be written,
M:N::P + A + A+C + mE:Q + B + B4-D-fmF,
without altering the truth of the proposition.
Thus, let 4 : 8 : B : 12, and 4 : 8 : : 6 : 10.
Then, 4 : 8 : : 6 + 5 : 12 + 10, and 4 : 8 : : 6 + 6 + 5 : 12+10+10.
And also, 4:8::6x9 + 6 + 5:12x9 + 10 +10.
Theorem XIV.
190. If four quantities are in proportion, the sum of the first ante-
cedent and consequent will be to their difference as the sum of the
second antecedent and consequent is to their difference.
For, since M : N : : P : Q, it follows that MQ = NP.
Add NQ to both members, then Q (M + N) = N (P + Q).
Subtract NQ from both members, then NXP — Q) = (M — N) Q.
Multiplying the two equations together, there results
(M + N)(P — Q) = (P + Q)(M-N)..
From which we get M + N : M — N : : P + Q : P — ^Q.
Let 2 ; 6 : : 8 : 24.
Then2 + 6:2 — 6: : 8 + 24 :8 — 24, or 8 : — 4 : : 32 :— 16,
9*
102 OE0M£TBI0AL PROPORTION.
Theorem XV.
191. If there are any number of proportional quantities, a single
proportion may be formed from them by multiplying together the cor-
responding terms of all the proportions.
For, let M : N : : P :
A : B : : C : D.
E : P : : G : H.
Then, MQ = NP.
AD = BC.
EH = FG.
Multiplying these equations, member by member, we get
(MAE) (QDH) = (NBF) (PCG).
Hence, MAE : NBF : : PCG : QDH.
Let 1 : 2 : : 4 : 8.
3 : 1 : : 9 : 3.
6 : 10 : : 2 : 4.
Then, also, 15 : 20 : : 72 : 96.
Theorem XVI.
192. If four quantities are in proportion, their like powers will be
proportional.
For, let M : N : : P : Q.
Then, MQ = NP.
Squaring both members, M'Q* = N'P*.
Hence, M» : N« : : P : Q».
In like manner it may be shown that M"" : N" : : P" : Q*.
Let 1 : 2 : r4 : 8.
Then, also, 1* : 2« : : 4^ : 8*, or 1 : 4 : : 16 : 64.
Likewise, 1» : 2' : : 4» : 8^ or 1 : 8 : : 64 : 51 2.
And similar proportions can be obtained for the 4th, 5th, &c.,
poweiB.
OSOMETBIOAL PROPOBTION. 103
General Remarks.
193. AH tlie foregoing theorems have beeu deduced from the first
two ; and it is obvious that an infinite series of proportions might be
deduced from them. The test of the truth of any proportion deduced
directly or indirectly from the first two theorems^ is the product of the
extremes being equal to the product of the means. Thus, if
M : N.: : P : Q; then, also, M + 4 : P + 4 ^ : N : Q.
For, by multiplying the extremes and means together, we have
MQ + 4Q = NP + 4Q,
which is a true equation when MQ = NP. In general, no propor-
tion is false, however absurd it may seem, when the product of the
extremes is equal to the product of the means. Thus, 1 : x* : : x~* : 1
is a true proportion, because the product of the extremes is equal to
unity, and that of the means also equal to unity.
The proportion M : N : : P : Q, gives MQ = NP, from which we
can get the four equations
M
P
N ''
Q'
M
P
N
Q
N'
P
Q
P
N
M'
involving eight distinct ratios. And, since ratio is the quotient arising
from dividing one quantity by another of the same kind, we conclude,
from the above scries of equations, that either the first consequent or
the second antecedent must represent quantities of the same kind as the
first antecedent; and that either the first antecedent or the second
consequent must represent quantities of the same kind as the first con-
sequent. We also see that two distinct species of quantities, and but
two, can be represented by a proportion. If two terms of a proportion
are abstract numbers, the proportion can represent but one kind of
quantity.
104 QEOMETRIGAL PROPORTION.
194. In the equation MQ = NP^ resulting from the proportion
M : N : : P : Q) if three terms are known, it is plain that the fourth
term can be found by solving the equation with reference to it. In
the Single Rule of Three of Arithmetic, three terms are given to find
the fourth. Thus, suppose the first three terms of a proportion are 3,
4, and 6. The fourth term can be found from the proportion 8:4::
6 : X, Hence 3a:; = 24, or x = 8, and the complete proportion is
3 : 4 : : 6 : 8. So, likewise, let the first three terms be a, 6, and r,
then the fourth results from the proportion, a \h \\ c\ x. Hence,
he '
a: = — . And we see that the fourth term can be found by multiply-
ing the second and third together, and dividing their product by the
first.
If the first, or either of the middle terms is unknown, we have only
to represent the unknown term by x, form the equation from the pro-
portion, and then solve it with reference to x.
Let the last three terms of a proportion be a, hy and c, to find the
first. We have, then, x : a : : & : c. Hence, d; = — .
c
So, we see that either extreme can be found by multiplying together
the means, and dividing their product by the other extreme.
Let the second term be unknown, and the other three be Hy hy and c.
oc
Then a \ x ; \h i c. Hence, x = -r-. Let the third term be unknown,
o
and the other three be a, hy and c. Then a i h \ i x : c. Hence, x =
J-, We see that either mean may be found by multiplying the ex-
tremes together, and dividing their product by the other mean.
EXAMPLES.
1. The first term of a proportion is a'; the third, c" j the fourth, m".
What is the second term ?
Ans. — =-.
<r
2. The first term of a proportion is a*"*^; the second, a~»; the
fourth, a"". What is the third term ? Ans, a*".
3. The three last terms of a proportion are a', 6", and (oft)'*. What
is the first term ? Ans. a*~»6~'.
GEOMETRICAL PROPORTION. 105
4. Given the first three terms of a proportion, a' — 6', a + 6, and
a — &, to find the fourth term. Ans. 1.
6. Given three first terms of a proportion, a', 6', and c, to find the
fourth. Arts, 6'c'a""'.
PROBLEMS IN GEOMETRICAL PROPORTION.
195. 1. Two numbers are to each other as 2 to 3; but, if 8 be
added to both, the sums thence arising will be to each other as 4 to 5.
What are the numbers ? Arts, 8 and 12.
Let X = smaller, then the greater will be |a;. For, 2 : 3 : : a; to
fourth term |a:. Then a; + 8 : |aj + 8 : : 4 : 5.
2. An author can write 240 pages in ten weeks : how many can he
write in 12} weeks? Ans, 300 pages.
3. Two quantities are to each other as a is to 5; but, if c be added
to both of them, the resulting sums will be to each other as ef to c.
What are the quantities ?
An,. ^'-^'^ and ^^T ^ ^.
od — ae ba — ae
4. An author can write a pages in h weeks : suppose .that he writes
uniformly, how many pages can he write in c weeks ?
. ac
Ans. a; = Y" P^g^s.
What values must be given to the letters in order to make Examples
3 and 4, the same as 1 and 2 ? How can the results in the last two
examples be verified ?
5. A father's age is now 3 times that of his son, but in 10 years
more the fiither's age will only be double that of the son. What are
their respective ages now ? Ans, Father, 30 ; son, 10.
6. A father's age is now a times greater than that of his son ; but
in h more years the father will only be c times older than his son.
What are the respective ages of father and son ?
Ans, Father, — ^^ ; son, ^: —,
a — c a — c
7. A gentU'fuan puts out his money at a certain rate of interest, and
derives 8500 income. He puts the same sum out a second year, one
per cent, more advantageously, and derives $600 from it. What are
the two rates of interest ? ^719. 5 and 6 per cent.
106 GEOMETRICAL PROPORTION.
8. The governor of a besieged town has provisions enough to aDow
each of the garrison 1} pounds of bread per day, for 60 days; but, as
he learns that succor cannot be expected for 80 days, he finds it neces-
sary to diminish the allowance. What must the allowance be ?
Ans. 11 lbs. per day.
9. A gentleman, who owns 20 slaves, had laid in a twelve-months'
supply for them, when he purchased 10 more. How long will his sup-
plies last ? Ans. 8 months.
10. A planter, who knows that his negro-man can do a piece of work
in 5 days, when the days are 12 hours long, asks how long it will take
him when the days are 15 hours long. Ans. 4 days.
11. Three negroes can hoe a field of cotton in 7 days : how long will
it take 4 to do the same work? Ans, bi days.
12. A family of 5 persons use a barrel of flour in 6 weeks : how
long will 2 barrels last 7 persons, using it at the same rate.
Ans. 8 1 weeks.
13. Two farmers purchase a piece of land for $1000 ; whereof the
first pays $600, and the second $400, and sell it again for $1200.
What proportion of the profit ought each to receive ?
Ans. The first, $120 ; the second, 180.
14. A bookseller purchased a certain number of books at the rate of
2 for a dollar, and also an equal number at the rate of 8 for a dollar,
and sold the whole at the rate of 5 for 3 dollars, and gained $22 by
the sale. What was the entire number of books he bought and sold ?
Ans. 120.
15. The hour and minute hands of a clock are together at 12 o'clock.
When will they be together again ?
Ans. 5-fj minutes past one o'clock.
Let X be the space after one passed over by the hour hand before
being overtaken by the minute hand. Then, 12 : 1 : : 60 -h x : as.
16. Two pedestrians start from the same point at the same time, to
walk around a race- course a mile in circumference. The first walks
11 yards in a minute, and the second 34 yards in 8 minutes. How
many times will the first have gone around the track before he is over-
taken by the second ? Ans. 33 times.
17. A brick-mason has brick 9 inches long, and 4} inches wide,
with which to build a wall 112 J feet long. He wishes to place 180
bricks in a row, and to lay some of them with their ends to the front
NEGATIVE QUANTITIES. 107
as heculers, and some of them lengthwise as ttretchers. How many
headers and how many stretchers most there he in each row?
Arw, 60 headers^ and 120 stretchers.
18. A hrick-mason wishes to place twice as many stretchers as
headers in a wall 105 feet long. How many of each kind must he use,
supposing the dimensions of the hrick the same as in the last prohlem ?
^718. 112 stretchers, and 56 headers.
19. A Freshman recited 5 times a week in mathematics, and his
average for the week was 66. His average for the first three days was
to the average for the last two as 7 to 6. What were those averages?
Ans. 70 and 60.
20. Three farmers purchase 900 acres of land for $9000 ; of which
the first pays $2000, the second $3000, and the tMrd $4000. What
share ought each to get, supposing that the land is equally valuable
throughout ?
Ans. The first, 200 acres ; the second, 300, and the third, 400.
21. A composition of copper and tin, containing 50 cubic inches,
was found to weigh 220-7 ounces. Assuming that a cubic inch of
copper weighs 4*66 ounces, and that a cubic inch of tin weighs 3*84
ounces, what must have been the relative proportion of tin and copper
in the composition ? Ans. Tin to the copper as 1 to 2i.
NEGATIVE QUANTITIES.
196. Quantiti^ are considered as negative when opposed, in charac-
ter or direction, to other quantities of the same kind that are assumed
to be positive.
If a ship, sailing at the rate of 10 miles per hour, encounter a head-
wind that drives it back at the rate of 8 miles per hour, then its rate
of advance will plainly be expressed by 10 — 8 miles per hour. Here
the retrograde movement, as opposed to the forward, is considered
negative. Suppose the wind to increase in violence until the ship is
carried back at the rate of 12 miles per hour. Then its rate of advance
will be expressed by 10 — 12, or — 2 miles per hour. The ship will
then plainly be carried back at the rate of 2 miles per hour, and we
see that the minus sign has indicated a change of direction.
108 NEGATIVE QUANTITIES.
If a man be now thirty years old. His age, four years hence, will
be expressed by 30 -h 4. Four years ago, it would have been ex-
pressed by 30 — 4. Here, future time being positive, past time is ne-
gative. And we see that a change of character is again indicated by
a change of sign.
If a man's money and estate bo represented by a, and his debts by
h ; then a — b will express what he is worth. His debts, as opposed
to his property, are considered negative. If h be greater than a, then
a — 6 is negative; and we commonly say that the man 'is worth less
than nothing. We see that there has been a change of sign in the
expression for what the man was worth, corresponding to a change in
the estimation of that worth. When the worth fell below zero, its ex-
pression became jpgative, because regarded as positive when above
zero.
If a man agree to labor for two dollars a day, and to forfeit one
dollar for every day that he is idle ; and he labor 4 days, and is idle 2 ;
then his wages will be 4 X 2 + 2 (—1), or 4 X 2 — 2 (+1) = 8— 2
dollars. We see that we have regarded as negative either the for-
feiture, as opposed to the gain, or the working days as opposed to the
idle.
Distance, regarded as positive, when estimated in one direction, must
be considered negative when estimated in a contraiy direction.
A B C
Let a distance, AB, estimated towards the right from the point A,
be represented by + m; and let a distance, BC, estimated towards
the right from the point B, be represented by -j- 7i.
Then, AC = AB -|- BC = m -|- n.
Now, suppose the point C be made to fall between A and B.
A C B
Then, AC = AB — BC = m — n.
And we see that the expression for BC has become negative, and
that this change of sign corresponds to a change in the direction of
BC. It was first estimated towards the right from B ; it is now esti-
mated towards the left from B.
NEGATIVE QUANTITIES. 109
Now, suppose BC be made greater than AB, then the point C will
fail on the left of A.
C A B_
And m — n, the expression for AC, will become negative. And we
see that the expression for AC, which was positive by hypothesis when
the distance was estimated on the right of A, has become negative
when the distance is estimated in a contrary direction.
197. The question now arises, &<« to what interpretation is to be put
upon a negative result when it appears in the solution of a problem.
Let it be required to find a number, which, when added to 6, will
give a result equal to 4. Then, from the conditions, we have the
statement 05 + 6 = 4. Hence, a; = — 2. Now, the problem is
purely arithmetical, and in arithmetic all numbers are regarded as posi-
tive. Therefore, the solution is absurd in an arithmetical sense, but it
is true in an algebraic sense. For substituting — 2 for x in the equa-
tion of the problem, we have — 2 + 6 = 4, a true equation. The
value satisfies the equation of the problem. It is, therefore, a true
answer to the problem in an algebraic sense. But it is not a true
answer to the problem as stated, for we were required to find a number,
which, when added to 6, would give a sum equal to 4 ; and we have
really found a number which, suhtracted from 6, gave a difference
equal to 4. The negative solution has then satisfied the equation of
the problem, but has failed to fulfil the conditions as enunciated. The
explanation of this difficulty is simple, when we return to the interpre-
tation put upon a negative quantity. We have said that a negative
quantity always indicates a change of direction or character. That
change must be marked by a corresponding change of condition in the
statement of a problem, and then the negative solution will be changed
into a positive one. The preceding problem must be changed into this :
required to find a number, which, when subtracted from 6, will give u
difference equal to 4.
Then, 6 — a; = 4, and ic = -f 2.
A negative solution will not, then, satisfy a problem as enunciated,
but will be the true answer to another problem in which there has been
a change of condition corresponding to the indicated change of charac-
ter. But negative solutions can be best explained by the discussion of
tte " Problem of the Couriers."
10
110 NEGATIVE QUANTITIES.
PROBLEM OF THE COURIERS.
198. Two couriers travel on the same road ; the forward courier at
the rate of h miles per hour, and the rear courier at the rate of a miles
per hour. At 12 o'clock they are separated by a distance of m miles.
The problem is, to determine how much time will elapse before they
are together, and also the point of meeting.
Let the indefinite line
A BO
represent the road on which they are travelling ; A, the position of the
rear courier ; B, of the forward courier at 12 o'clock ; 0, the unknown
point of meeting, and x the required time of meeting. Then, from the
conditions of the problem, aa; = AO, fea; = BO, and AB=rm; and
since, from the figure, we have, AO — BO = AB = m, we get the
statement, ax — hx = m. Hence, x = r. The distance thev
are apart at 12 o'clock, divided by the difference of their rates of travel,
will give the number of hours that must elapse^ after 12, before they
are together. The time, multiplied by the rate of travel, a, will give
th^ distance, AO, from A to the unknown point of meeting' ; or the
time, multiplied by the rate b, will give the distance, BO. If, for in-
stance, they are sepiurated by a distance of 48 miles at 12 o'clock, and
the forward courier travels at the rate of 4 miles per hour, and the rear
courier at the rate of 6 miles per hour, then, m = 48, a = 6, and
6 = 4. Hence, x = ;; 7 = 24 hours, ax = 6 . 24 = 144 miles
b — 4
= AO, ftx = 4 . 24 = 96 miles = BO. Verification, AB 4- BO
= 48 + 96 = 144 = AO.
We might make AO the unknown quantity, and call it y ; then, BO
= V — m. Then, — , the distance the rear courier will have to travel,
a
divided by his rate of travel, will give the time that must elapse before
he will overtake the forward courier. So, ^—r — will give the time that
the rear courier will be travelling before he is overtaken. And, since
these are but different expressions for the same time, we get the equa-
y — m y „ am , . , - ,
tion '^— 7 — = — . Hence, y = . ; and supposing, as before, that
NEGATIVE QUANTITIES. Ill
a = 6, 2) = 4. and m = 48, we have v = AO = rr- — r = 144 miles.
•^ o — 4
y 144
Then, — =3 -^- = 24 hours, as before.
a o
199. We will confine our discussion mainly to the expression, x sa
m
-, in which x represents the unknown time. Since the value of
a — 6
a fraction increases as its denominator decreases, it is evident that, by
making the difference between a and b very small, we can make the
time very great. This is apparent, too, from the problem ; for, if the
rear courier travel but little faster than the forward, he will gain but
little upon him. Now, if the denominator be made the smallest possi-
ble, the fraction will be the greatest possible. Hence, when the deno-
minator is zero, which results from making h =i a, the fraction must
have the greatest possible value ; or, in other words, an infinite value.
Therefore, x =s — = infinity, symbolized by the character 00 . It will
then be an infinite time after 12 o'clock before the couriers will be to-
gether; that is, they will never be together. This is plain, from the
nature of the problem ; for, if the couriers are separated by a distance
of m miles at 12 o'clock, and travel at equal rates after 12, they must
always keep at the same distance of m miles apart.
The character 00 , then, indicates impossibility. Whenever it ap-
pears, by examining the equation of the problem, we will discover an
absurdity arising from the hypothesis that gave the infinite result. In
this case, infinity was the result of the hypothesds, a = 2^. When
a == &, ox is equal to Z»x, or AO = BO : a part equal to the whole,
which is absurd. ^
The character 00 , then, indicates two distinct things, viz. : impossi-
bility in the fulfilment of the required conditions, and absurdity in the
conditions themselves.
200. Now, let h become greater than a, the denominator of the value
of X becomes negative, and the whole fraction negative. What does
this solution mean ?
The sign of the time being minus, indicates that it refers to past
time, because in the statement we regarded future time as positive.
It is plain, also, from the nature of the problem, that the time of the
couriers' being together, if together at all, must be past, not future.
For the forward courier travelling more rapidly than the rear courier,
Il2 NEGATIVE QUANTITIES.
will never be overtaken by him. The solution, then, fails to apply to
the problem as enunciated ^ but it is a true solution for another ques-
tion, viz. : how long before 12 o'clock were the two couriers together?
Let us return to the particular problem : making & ac 6 instead of
tn
a = 6. Then, m = 48, a = 4, ft = 6, and x = r = — 24.
a — o
Now, it is obvious, that the forward courier, gaining two miles per hour
on the rear courier, must have been with him 24 hours before 12,
otherwise the two would not be separated by 48 miles at 12.
CLfTV
Returning to the equation in distance, y = r, we see that the
distance, AO, also becomes negative under the hypothesis of a <[ 5.
This solution indicates impossibility for a point in advance of B, be-
cause, when a <^bf ax is also <^ bxy or AO <^ BO, which is absurd.
But it will be a true solution for a point in rear of A.
I
A B ^
^ I
Because, for such a point, the relation ax <^ hx will be true.
Resuming the values m = 48, a = 4, and ft = 6, we get y =: AO
= — 96 miles ; which agrees with the solution above, for the courier
A, 24 hours before 12 o'clock, being at 0, must be 96 miles fo>m
at 12.
When the unknown quantity was time, we have seen that the n^a-
tive solution indicated impossibility for future time, but gave a true
answer to the question in past time. When the unknown quantity was
distance, the negative solution indicated impossibility for forward dis-
tance, but gave a true answer to the question in distance estimated in
the opposite direction. In general, a negative solution indicates that
the problem, as enunciated, cannot be solved, but gives a true solution
when the unknown quantity is changed in character or direction.
That being the case, a negative solution may always be changed into
a positive solution by changing the character or direction of the quan-
tity that produced the negative result. The negative solution, x =
AVI
r , may be changed into a positive solution in three ways. First,
by changing the direction of the time ; that is, by changing + x into
— X, which is equivalent to asking, how long before 12 o'clock the two
couriers were together. Then the equation, ax — hx = m, becomes
NEGATIVE QUANTITIES. 113
Ix — ax = m, or X = r , a positive result, because b is greater
than a. Second, by cbanging — h into + b, which is equivalent to
turning the forward courier around to meet the rear courier. The
AM
equation then becomes x = — -—7. The diagram corresponds to this;
* a -{• o
for, in this case, the point will be between A and B, and we will
have ax + 6x = AG + BO = AB.
A B
Third, by changing + a into — a, and — b into + b, which is equi-
valent to turning both couriers around, and making the courier B go in
pursuit of the courier A, the equation becomes — ax + bx ss m, or
X = = , a positive result. It will be seen that the results in the
— a
first and third cases are the same, as they obviously aught to be.
201. Now, let m = 0, and a > b. Then, x = -=- = 0. The
a — b
quotient of zero by any finite quantity is plainly zero ; for the numera-
tor of a fraction indicates the amount to be divided, and the division
of nothing must give nothing. Thus, if a man has zero dollars to di-
vide among three persons, the share of each will plainly be nothing.
In the present instance, the solution, 0, shows that the couriers will be
together at a zero time after 12 ; in other words, at 12 itself This
ought to be so; for, when m = 0, they are together at 12 ; and, since
they travel at unequal rates, they will be together no more.
202. Now, let »i = 0, and a = 5. Then, x = -:-. To explain this
symbol, it is necessary to notice particularly the conditions. The con-
dition, m = 0, places the two couriers together ; the condition, a = &,
makes them travel at equal rates. Being together at 12, and travelling
at equal rates, they must always be together. They will then be to-
gether at 1, at 2, 3, 3}, &c. There is, then, no determinate time at
which they are together. Hence, — is called the symhol of indetermt-
nation. It does not indicate that no solution can be found, but, on the
contrary, that too many can be determined ; and the indetermination
consists in this, that any one of the infinite solutions will answer just
10* H
114 NEGATIVE QUANTITIES.
as well as any other. Suppose, for instance, the answer to the qaes-
tion, who discovered America? was, an inhabitant of Europe some
time after the Christian era. The answer would be indeterminate, be-
cause equally applicable to countless millions.
By clearing the equation, x = -rr, of fractions, we get . x = 0.
This will be a true equation, when a; = 1, 10, 1000, anything what-
ever. Hence, we say that x is indeterminate. The diagram, also,
shows the same thing.
A
B
For, when m = 0, the points A and B become the same ; and, since
a = hf then, AO = BO in all positions of the point 0.
Thus, it is shown in three ways that jr is the symbol of indetermi-
nation. The symbol also arises from an identical equation ; for, when
a = 5, and m = 0, we have ax = ax, an identical equation, in which
X may have any value whatever.
By recurring to the equation in y, we will observe that the distance,
from the common point of starting to the unknown point of meeting,
also becomes indeterminate when m = and a = 6. This, obviously,
ought to be so.
203. There is one remarkable exception to the foregoing statement^
that -rr- is the symbol of indetermination. It is in the case of vanishing
fractions. A vanishing fraction is one, which becomes -r-, in conse-
quence of the existence of a common factor to the numerator and deno-
minator, which has become zero by a particular hypothesis made upon it
204. Take the expression - -rzT^ = 7r> ^^^^ m = n, but, by de-
composing the denominator into its factors, we have
(m — n) (m 4- «)"
We see that -rr is caused by the common factor, m — w, becoming
by the hypothesis m = n. Divide out this factor, and we have left
— ; — = — , when m ^ n,
m -\- n 2n
NEGATIVE QUANTITIES. 115
Agam^ take the fraction^ = ^^ — = -jr-, when
m = n. But, divide out m — n, and we have — = — = 2ft, when
«, (m — nf (m — n) (m — n) . _
m =2 n. So, ^ = ^ ^-^ = TTi when m = n. But
TO ft TO — ft
by division, the fraction becomes — = — = 0, when m =: n. So,
TO ft TO ft ^1. TJxi.'i*'
= -^, when TO = n. But, by divi-
(to — »)* (to — n) (m — n)
sion, we get = ~ = oo, when to = n.
' ^ TO — ft
We conclude, that a vanishing fraction has one of three true values;
that it may be either finite, zero, or infinity. How, then, are we to
decide whether the symbol jr indicates indetermination, or a vanishing
fraction ?
If the particular hypothesis which gives the symbol, does not make
the given equation an identical equation, we may be certain that j-
points out a vanishing fraction.
We will resume the subject of vanishing fractions more at length
hereafter. '
205. The foregoing symbols are of the highest importance, and ought
to be remembered.
Arrangement in tabular form will assist the memory.
TO
X = -:r- = 00!
'
that is, a finite quantity, divided by zero, equal to infinity. The sym-
bol, oo, indicates impossibility in the fulfilment, and absurdity in the
conditions of the problem.
that is, zero, divided by any finite quantity, equal to zero. This is a
true solution, unless it conflict with the condition of the problem.
5C = — A.
The negative solution indicates that the problem, as enunciated, can-
c
116 GENERAL PROBLEMS.
not be solved, but the negative resnlt will be a tme answer when tbe
unknown quantity is changed in character or direction.
_
^- p
the symbol of indetermination, when there is no common factor in
the numerator and denominator. By indetermination is meant, that
there is an infinite system of values that will satisfy the conditions of
the problem.
GENERAL PROBLEMS.
206. 1. A person employed a workman for m days, upon conditiou
that the workman should receive n dollars every day that he worked,
and should forfeit jp dollars every day that he was idle ; at the end of
the time he received c dollars. How many days did he work, and how
many was he idle ?
Let X = working days, then m — x = idle days ; and by the con-
ditions, we get nx — (m — aj) j) = c.
TT c + mp .ji J c-\-mp nm — c
Hence, x = -, m — x = idle days = m — —
n -\- p "^ n -^ p n -{- p
When will the last expression be zero, and when negative, and what
do these solutions indicate ?
The negative solution needs some explanation : c, the entire amount
received, cannot exceed nm, what was paid for working m days, unless
the workman had not only no forfeiture to pay, but also worked more
than the time for which he was employed. The idle days have then
changed their character, and become working days beyond the period
for which the man's services were engaged.
2. The same problem as before, except the workman, at the end of
the time, was in debt c dollars.
. mp — c , tim 4- c
Ans, X r= -^ and m — x = .
w + m « + p
When will the first solution become zero, and when negative ?
^ 3. The same problem, except that the workman received nothing.
. TKip . ntn
Afu, X c= — ±— and m — a: = .
n -\' p '* + p
The above results are formulas, and may be made applicable to par-
ticular problems of the same kind.
4. In a certain college, the maximum mark for a recitation is 100.
OXNERAIi PE0BLEM8. 117
A Freshman recited 5 times in mathematios; two days he got 95
for each recitation, and his averago for the 5 days was 86. What did
he average each of the other three days t Ans. 80.
5. Divide the number a into two such parts that the product of the
first by m shall be equal to the quotient arising from dividing the
second by n.
Ans. =— , and
1 + nm 1 + nm
What effect has increasing n upon both results ? Under what form
must the second expression be placed to show that it is directly propor-
tional to n and m/ When will the two expressions be equal ?
6. A father is 32 years old, and his son 8 years. How long will it
be until the age of the father is just double that of the son ?
Ans. 16 years.
7. A father is a years, and his son h years old. How long until the
age of the father will be c times as great as that of the son ?
a — ch
Ans. X ^ =-.
c — 1
What do these values become when c = 1 ? What, when ch^^ a?
What, when ch^al What do these different solutions indicate ?
When c •= 1, x = oo, the symbol of absurdity, as it ought to be.
The hypothesis makes the father and son equal in age after the lapse
of h years. When c6 = a, x = 0. A true solution, since the father
is now c times as old as his son. When c5 ^ a, x is negative. Fu-
ture time being positive, past time is negative. The solution, then,
indicates that =- years ago, the father was c times as old as his
8on.
8. A father is 64, and his son 16. How long will it be until the
age of the father is 9 times as great as that of the son.
Ans. X ^ — 10.
The solution indicates that 10 years ago the age of the father was 9
times as great as that of the son. The father was then 54, and the
son 6 years old.
9. Bronze cannon (commonly, but improperly, called brass cannon)
are composed of 90 parts of copper, and 10 parts of tin : S^^y lbs. of
copper lose 1 lb. when immersed in water, and 7 j^ lbs. of tin lose 1 lb.
when immersed in water. The Ordnance Board suspecting that some
bronze cannon did not c;.)ntain copper enough, immersed one of them,
118 GENEBAL PROBLEMS.
weighing 900 lbs., and found its loss of weight to be lOSJff j^ lbs.
What was its composition ?
Ans. 80 parts of copper, and 10 parts of tin.
Verify the results. The copper lost 89|| lbs., and the tin IS^^ lbs.
The sum of which is 103||$f lbs.
10. A fox, pursued by a greyhound, is 125 of his own leaps ahead
of the greyhound, and makes 6 leaps to the greyhound's 5, but 2 of
the greyhound's leaps are equal to 3 leaps of the fox. How many leaps
will the fox make before he is caught by the greyhound ?
Let X as number of leaps made by the fox ; then, since the grey-
hound makes but 5 leaps while the fox makes 6, he will make | cf a
leap to one leap of the fox. Therefore, he will make |x leaps whOe
the fox makes x leaps. But each of the greyhound's leaps are equal
to I of a leap of "the fox. Hence, the §x leaps of the greyhound are
equivalent to | . |x leaps of the fox ; and since the greyhound has not
only to run over the ground passed by the fox in making x leaps^ but
also that pa^ed in making 125 leaps, we have the equation of the pro-
blem. I . |x =s a; -f 125. Ans. x = 500 leaps.
Verification. While the fox made 500 leaps, the greyhound made
416} leaps ; and these were equal to | . 416 J, or 625 leaps of the fox,
the entire number of leaps made by the fox.
Eemarks,
This problem shows the importance of making the two members ex-
press the same thing by referring them to the same unit. The leaps
of the greyhound have been expressed in terms of those of the fox.
We might have made the unknown quantity represent the number of
leaps made by the greyhound, and, in that case, the leaps of the fox
must have been expressed in terms of those of the greyhound.
11. A fox, pursued by a greyhound, has a start of a leaps, and
makes h leaps while the greyhound makes c leaps ; but d leaps of the
greyhound are equal to e leaps of the fox. How many leaps will the
fox make before he is overtaken by the greyhound ?
. adb
Ans. X = ^T.
cc — do
When will x^of When will it be equal to infinity ? When wiU it
be negative ? How are these solutions explained ?
X will be zero, when a is zero, x will be infinite when ec^^^dh;
GENERAL PROBLEMS. 119
that is, when the number of leaps made by the greyhound, multiplied
by the value of each leap, is equal to the number of leaps of the fox,
multiplied by the value of each of his leaps. In that case, the hound
will evidently never overtake the fox ; and the solution indicates impos-
sibility or absurdity, x will be negative when dh is greater than ec;
then the fox is running faster than the hound, and the distance re-
presented by the x leaps must be estimated in a contrary direction.
F H F V_
Let H be the position of the hound ; F, that of the fox ; and P the
point where the fox is overtaken by the hound. Then, when x is ne-
gative, we understand either, that the fox pursued the hound and
caught up to him at some point, P' on the left^ or that, at some time
previous to the fox being at F, and the hound at H, they were both
together at P', and the fox running faster than his pursuer gained upon
him the distance HF = a.
12. A man, desirous of giving 4 cents apiece to some beggars, found
that he had not money enough by 5 cents ; he therefore gave them 2
cents apiece, and had 15 cents left. How many beggars were there,
and how much money had he ? An$, 10 beggars : 35 cents.
13. A man, desirous of giving a cents apiece to some beggars, found
that he had not money enough by h cents ; he therefore gave them c
cents apiece, and had d cents left. How many beggars were there,
and how much money had the man ?
. d + h . ^ ad 4- ch
Ans, bepcars, and cents.
a — c a — c
What values must a, 6, c, and d have to make this problem the
same as the last ?
When c ^ a, the solution is negative ; and negative solutions indi-
cate a change of direction and character. The beggars become givers :
the amount given becomes the amount received : the deficiency be-
comes a surplus, and the surplus a deficiency. When c = a, both so-
lutions become infinite, and the symbol, a, here plainly indicates
absurdity.
When d and h are both zero, the two solutions will be both 0, or
F "^^^
14. A man, desirous of giving 2 cents apiece to some beggars, found
that he had not money enough by 15 cents ; he therefore gave them
120 OENEBAL rROBLEMS.
4 cents apiece, and had 5 cents left. How many beggars were there,
and how much money had the man ?
Ans» — 10 beggars, and — 35 cents.
How is the solution explained ? There were 10 givers, who each
gave the man 2 cents, and 15 cents over; or each 4 cents apiece, lack-
ing 5 cents in all.
15. A piratical vessel sails at the rate of r miles per hour for a
hours, when she sustains some injury, and can only sail i^ miles per
hour. At the moment in which the piratical vessel is disabled, a
sloop-of-war starts in pursuit, and sails at the rate of r miles per hour,
from the point where the pirate fii-st started. How long before the
sloop will overtake the pirate ?
Ans. X = ; hours.
r — r
When will this solution be zero ? When negative ? When infinite ?
Under what form must the fraction be placed to show that x is recipro-
cally proportional to r ^
Make the general solution applicable to a particular example.
16. Divide a number, a, into two parts, which shall be to each other
as m is to n.
. ma _ va
Ans. , and
m + n m 4- »
17. Divide a number, a, into three parts ; such, that the first shall be
to the second as n to m, and the second to the third as (^ to p.
. a7iq amq amp
mp -h mq + "2' '^P "^ ^? + ^?' '^^P + ^'2 "^ **?
Wliat supposition will make all the parts zero ? What one of them ?
What two of them ?
18. Divide a number, a, into four parts ; such, that the first shall
be to the second as n to m, the second to the third as j to j9, and the
third to the fourth as r to a.
. anqr amqr
Ans, , ,
nqr + mqr 4- mpr -\- mps nqr + mqr -f wpr -f- mps
ampr amps
nqr + mqr -\- mpr -f mps^ nqr -f mgr + mpr 4- mp '
19. Milk sells in the City of New York at 4 cents per quart. A
milkman mixed some water with 50 gallons of milk, and sold the mix-
ture at 3 cents per quart without sustaining any loss by the sale. How
much water did he pat in the mijk ? Ans. GC3 quarts.
QENERAL PBOBLSMS. 121
20. Milk sells in Boston at a cents per quart. A milkman mixed a
certain quantity of water with b quarts of milk, and sold the whole at
r cents per quart without losing anything by the sale. How much
water was added to the milk ?
ah — he
Ans. X = quarts.
c
The value of x is zero when c:=-a. In that case^ evidently, no
water is added. The value is infinite when c =s o. Then the milkman
gives away, gratuitously, an infinite quantity of water. When c ]> 6,
X is negative. Then we understand that a certain quantity of water is
separated from, not added to the milk, and that the price of the milk is
r cents per quart, and of the mixture a cents per quart.
21. How often are the hour and minute hands of a clock together?
Anz, Every 65^j minutes.
22. How often are the minute and second hands of a clock to-
gether? Am, Every 1^^ minutes.
23. How often are the hour, minute, and second hands of a clock
together '( An%, Every 720 minutes.
The last problem is solved by means of the least common multiple.
24. There is an island 60 miles in circumference. Three persons
start from the same point to travel around it, travelling at the respect-
ive rates of 4, 6, and 16 miles per hour. How often will all three be
together ? Am, Every 30 hours.
This problem is solved by means of the least common multiple.
25. There is an island a miles in circumference, around which three
persons start to travel, at the rates of 2>, c, and d miles per hour.
When will they all be together again ?
Ant. In a hours, divided by the greatest common divisor of c — 6,
and d — c.
26. Same problem as 24, except that there are 4 persons travelling,
at the respective rates of 3, 6, 12, and 27 miles per hour.
Anz, Every 20 hours.
27. A former purchases a tract of land for $500, on a credit of 10
months, or $480 cash. What is the rate of interest that makes these
sums equivalent. Ant, 5 per cent.
Let X = interest upon $100 for one month. The statement will bo
100 -h lOx : 100 : :' 500 : 480.
11
122 QENEBAL PROBLEMS.
28. A fiurmer buys a tract 6f land for a dollars, payable in h months,
or for c dollars cash. What is the rate of interest ?
100 (a — c)
Ans, X s=
ch
What supposition will make this yalue zero ? What two sapposi-
tions will make it infinite ? What supposition will make it negative ?
and how is the negative solution explained ?
29. Two numbers are to each other as 8 to 3 ; but if 8 be added to
both numbers, the first will only be double the second. What are the
numbers? Ans, 32 and 12.
30. Two numbers are to each other as a to Z) ; but if c be added to
both of them, the first will only be d times as great as the second.
What are the numbers ?
'A w ^ ac (d — 1) ^ ^c (d — 1)
Ans. First, — ^^ — z- .— : second — ^= — r-r^-
a — oa a — bd
What do these solutions become when cf = 1 ? What, when
hd sss al What, when bd'^ a? What, when c? s= 1, and bd =: al
The first hypothesis gives a true solution. The second gives an ab-
surd solution, as it ought, since bd can only equal a when the first
number, after the addition of c to both numbers, exceeds the second
proportionally as much as before. But this is impossible, since the
smaller increases most rapidly. See Article 96.
The hypothesis, bd ]|> a, gives a solution impossible for arithmetical
quantities, but possible for algebraic. We must either suppose that
two numbers are to be found, which result from the subtraction from a
number not expressed, or we must change the character of the problem,
and make c subtractive. ' When bd = a, and <^ = 1, the two numbers
are represented by J, the symbol of indcterminatiou. An infinite, or
indeterminate, number of quantities will satisfy the conditions of the
problem.
The following problems will illustrate the foregoing cases :
31. Two numbers are to each other as 4 to 3 ; but when 5 is added
to both, the numbers are equal. That is, d = 1.
Ans. Both numbers zero.
32. The same as last, except that, after the addition, the first will be
I greater than the second. That is, hd = a.
Ans, Both infinite.
QSMERAL PB0BLEM6. 123
83. The same as 31, except that, after the additions, the first will be
twice as great as the second. That is, hd '^ a.
Ans, — 10, and — 7i.
Change the character of the problem, and take 5 from both numbers,
and the solutions will be + 10 and + 7.
34. The same as 31, except that the numbers were equal before the
addition of 5 to both.
Ans. Both indeterminate, -jr. Any numbers will answer.
35. A father divided his estate, worth $1200, among his three sons,
so that the share of the first should be to that of the second as 6
to 4 ; and so. that the share of the third should be the greatest common
divisor of the shares of the first and second. What will be the share
of each ? Ans. $600, $400, and $200.
■ ■
36. A father divides his estate among his three sons, so that the
share of the first shall be to that of the second as — - is to -— : and so
that the share of the third shall be the greatest common divisor of the
other two. The father's estate is worth a dollars : what is the share
of each son ?
Ans. First, : =- : second, := : third, :, .
ms -\- nq -\- 1 w« -f nj -f 1 wi« -f 7ij + 1
When will the shares of the first and second be equal ? What will
the share of the third be then ? What will be the effect upon the
np
share of the first to make — ^ =s oo ?
s
37. The difference of 'two numbers is a, and the difference of their
squares is zero. What are the numbers ?
. a a
38. The sum of two numbers is 2a, and the difference of their
squares equal to c. What are the numbers ?
. 4a' 4- c - A:d^ — c
Ans. — -. , and — ; .
4a 4a
What do these solutions become when c = ? What, when
c = 4a2 ? What, when c > 4a' ?
124
GENERAL PROBLEMS.
39. Two ships sail in such a manner that the track of one cute the
parallel of latitude through the Island of St.
Helena 40 miles on the west of that island,
and cuts the meridian through the island 40
miles north of it; and that the track of the
other cuts the parallel of latitude 80 miles on
the east, and the meridian 80 miles on the
north? Where will the two tracks intersect
each other ?
Ans. 20 miles east, and 60 miles north of the island.
Call DO, X ; then, PD = a- + 40 ; and PD = 80 — x. Hence,
40 + ic = 80 — a-; then, x = 20 ; and PD = 60.
40. A Yankee mixes a certain number of wooden nutmegs, which
cost him \ cent apiece, with a quantity of real nutmegs, worth 4 cents
apiece, and sells the whole assortment for 844 ; and gains $8*75 by the
fraud. How many .wooden nutmegs were there?
Ans. 100.
41. A Yankee mixed a certain quantity of wooden nutmegs^ which
cost him -=-— part of a cent apiece, with real nutmegs, worth c cents
apiece, and sold the whole for a dollars. He gained d cents by the
ftnud. How many wooden nutmegs were there ?
dh
Ans. X := r .
DC 1
What does this become when c? = ? What, when & = ? What,
when Z»c = 1 ? What, when he <C^1,
42. The sum of three numbers is 200 ; the first is to the second as
5 to 4, and the third is the greatest common divisor of the first two.
What are the numbers ? Ans. 100, 80, and 20.
43. The sum of three numbers is a ; the first is to the second as h
to c, and the third is the least common multiple of the first two. What
are the numbers ?
. ah ac ahc
h + c + hc' h + c + hc' h + c + hc'
What single hypothesis will make them all equal ? What will these
values become when o = 115, 6 = 25, and c = 15.
41. A northern railroad company is assessed $120,000 damages
for contusions and broken limbs, caused by a collision of cars. They
GENERAL PROBLEMS. 125
pay $5000 for each conttLBion, and $6000 for each broken limb ; and
the entire amount paid for braises and fractures is the same. How
many persons received contusions, and how many had their limbs
broken 1 Ans. 12 of the former, and 10 of the latter.
45. Same problem as the last, except representing the assessment by
a, the price of each contusion by c, and that of each broken limb
by 6.
What do these values become when c = ? What, when 6 = f
What, when a = ? What, when h ==: c?
46. The reservoir at Lexington contains 48,000 gallons of water,
and supplies the town and Virginia Military Institute. If all the con-
ducting pipes were closed, the reservoir would supply the town and
the Institute for 57{ hours, and the town alone for 96 hours. How
many gallons does the Institute use per hour.
Ans. 333 i gallons.
47. Two pipes will exhaust a cistern containing a quantity of water,
represented by q, in a hours, and the first will, alone, exhaust it in />
hours. How long will it take the second pipe to empty it, and how
much does it exhaust per hour.
Ans, T hours, and — — = — - gallons per hour.
o — a ab
What do these solutions become when a = &, and a^ h?
The negative solution can be illustrated by an example.
48. Two pipes, a and 6, will exhaust a cistern in 4 hours ; and the
pipe a can, alone, exhaust it in 2 hours. In what time can the pipe h
empty the cistern ? Ans. In — 2 hours.
The solution being negative, may refer to past time, and indicate
that, during the two hours before the pipe a began to play, the pipe b
had exhausted the cistern. The two hours play of the pipe &, added
to the two of the pipe a, give the 4 hours of the problem. Or, we
may suppose that the character of the pipe b has been changed, and
that it has been a supplying pipe for two hours, and, consequently, the
pipe a has been twice the time in exhausting the cistern.
49. A man asks $120 cash for his horse, or $126*30 on a credit of
9 months. Supposing these valuations to be equal, what is the rate of
interest ? Ans. 7 per cent.
11*
126 GENERAL PROBLEMS.
50. A man yalues his horse at c dollars cash, or b dollars on a credit
of a months. What is the rate of interest ?
Aru .^- 1200(6-c)
ac
What does x become when h = cf What, when a^hf What,
when a, or c = ?
51. There is an island 82 miles in circamference : two persons start
to travel around it, the first at the rate of 11 miles per day, and the
second at the rate of 3 miles per day. When will the distance be-
tween them be equal, estimated in either direction around the island ?
Alls. At the end of the second day.
52. Two persons start to travel around an island a miles in circum-
ference, the first at the rate of h miles per day, and the second at the
rate of c miles per day. When will the distances between them be
equal, estimated in both directions around the island ?
Aru* X =5 - —
2 (b — c)
What will this solution become if the travellers move at equal rates ?
What, if the second travels the fastest ? What, if the second stops ?
53. Three persons start to travel around an island which is a perfect
circle, 120 miles in circumference. The first travels at the rate of 4 miles
per day, the second at the rate of 8 miles per day, and the third at the
rate of 6 miles per day. How many days will elapse until the lines
joining their respective positions, and the centre of the island, will
trisect the circumference, under the supposition that the second, only,
has gone past the starting point on a second tour of the island ?
Ans. 20 days.
54. Three persons travel around a circular island, 160 miles in cir-
cumference. They start together, and travel at the respective rates of
7, 5, and 2 miles per hour. How long will it be until the third will
be between the first and second, and equally distant from them, under
the hypothesis that the first and second, only, have passed the starting
point ? Ans. 40 hours.
55. Three persons travel around an island a miles in circumference.
They travel at the respective rates of i, c, and d miles per hour. How
long will it be until the third will be half way between the first and
second, and how fiir will the first have travelled ?
Ans, Distance, 7 ^-r— — ; time.
h — 2d + c' 'b — 2d + c
GENERAL PROBLEMS. 127
What do these Talnes become when 2d sssb + cf What, when
2d ss cf How do you explain these results ?
56. Divide the number 24 into two such parts that their product
be the greatest possible.
Let X express the excess of one of the parts over the half of 12,
then, 6 + ^9 ^^^ 6 — x will represent the two parts. And (6 + x)
(6 — x), 36 — x* is to be the greatest possible. This result will evi-
dently be the greatest possible when x ass 0, or when the parts are
equal.
57. Divide the quantity a into two such parts that their product
shall bo the greatest possible. What are the parts ?
Ans. — , aud —.
58. Two men have each an end of a pole 6 feet long upon their
shoulders, with a burden suspended from it. The share of the load sus-
tained by each man is inversely proportional to the distance of the load
from his shoulder. The proportion of the weight borne by one man is
to that borne by the other as 3 to 2. How far is the burden from the
shoulder of the first man f Ana. 2| feet.
59. Same problem as the last, except the representation of the length
of the pole by a, and the propdHion by h and c. How far is the bur-
den from the shoulder of each man ?
Am. ' — -— feet, and z — : — feet.
+ c b + c
When will these distances be equal ? When one double the other ?
60. A general wishing to range his men in a solid square, found
that he had too many men by 100. He increased each side of the
square by one man, and then found that he had but 39 too many
How many men had he ? Ans. 1000.
Let'x =3 side of the square.
61. A general ranged his army in a solid square, and found that he
had a more men than would enter the square. He then increased
each side of the square by one man, and found that he had b more
men than could enter the square. How many men were in each side
of the square ?
Am. X = K: L
128 GENERAL PaOBIiEHS.
What does this value become when b + 1 tss at What, wlien
is — 1 ? How do you explain these results ? The last result will
confirm a truth hereafter to be demonstrated.
62. Divide the number 20 into three such parts that the half of the
first, the one-third of the second, and the one fifth of the third shall be
equal. What are the parts ? Ans. 4, 6, and 10.
For, let X represent the equal result after the division of the three
parts by 2, 3, and 5 ; the parts themselves will plainly be represented
by 2x, 3x, and 5a-. Hence, 2x -f 3x + 5x = 20, or a: = 2. Then,
2x = 4, 3x = 6, and 5x =: 10.
63. Divide the number a into three such parts that the first divided
by by the second by c, and the third by d, shall all be equal. What
are the parts ?
. ab ac , ad
Ans. •= — : : — ,. z — : : — ;, and
h^c + d' b + c + d' b + c + d'
When will the three parts be equal ? What will be the effect of
making 6 = 0? How do you explain the result ?
64. Two laborers are engaged to dig a ditch ; the first can dig it in
5 days, the second in 3 J days. How long will it take the two, working
together, to dig the ditch ? A.n$. 2 days.
Verify the result.
65. Two men are employed to dig a ditch ; the first, alone, can dig
it in a days, and the second, alone, can dig it in b days. How long
will it take the two, laboring together, to perform the work ?
ab
Ans.
X =
6 + a
When will the time be one half of that required by the first to dig
it ? What change will this solution undergo when the second man fills
up instead of digs out ? What will be the expression for the time in
that case, when he fills up as fast as the first digs out ?
This problem shows clearly that a change in a condition will be
accompanied by a change of sign.
66. A debt of 8150 was paid in dollar and five-cent pieces. The
dollar and five-cent pieces were 1100 in number. How many dollar,
and how many five-cent pieces were used ?
Ans. 100 dollar pieces, and 1000 five-cent pieces.
67. A debt of a cents waa paid in b and c cent pieces, and the total
SLIMINATION BBTWI£N XQUATIONS. 129
number of these pieces was equal to df How many pieces of each
land were used.
. hd — a J a — cd
Ans, -T , ana -= .
6 — c b — c
What do these solntions become when hd^^ a, and cd ^s al What,
when a > ft^, and cd"^ af What, when h^cf Explain these
rcsnlts, especially the last.
68. The sum of three numbers is 420 : the first is to the second as
6 to 7, and the third is just as much greater than the second as the
second is greatei* than the first. What are the numbers?
Ans, 120, 140; and 160.
69. The sum of three numbers is a : the first is to the second as
& to c, and the second exceeds the first as much as the third exceeds
the second. What are the parts ?
. ac a ^ 2ah — ac
^"* Wl^^ 36 •
We observe that, when quantities bear the above relation, the mean
is always the third of the whole.
What single hypothesis will make all three parts equal ? What is
the effect upon the three parts of making c aa 25 ? Explain this
result?
70. At the Woman's Rights Convention, held at Syracuse, New
York, composed of 150 delegates, the old maids, childless-wives, and
bedlamites were to each other as the numbers 5, 7, and 3. How many
were there of each class ? Ans, 50, 70, and 30.
ELIMINATION BETWEEN SIMULTANEOUS
EQUATIONS OF THE FIRST DEGREE.
207. Simultaneous equations are those which can be satisfied for the
same values of the same unknown quantities. Elimination can only be
performed upon simultaneous equations. It is a process by which we
deduce from two or more equations, containing two or more unknown
quantities a single equation, containing one unknown quantity. The
single equation so found is called the final equation, and since it con-
tains but one unknown quantity does not lead to arbitrary values.
I
180 ELIMINATION BETWEEN SIMULTANEOUS
We wili first take the simplest ease^ that of two equatioos inydting
two unknown quantities. The elimination of one of these unknown
quantitios may be effected in four wajB.
1. By Comparison.
2. By Addition and Subtraction.
3. By Substitution.
4. By the Greatest Common Divisor.
ELIMINATION BY COMPARISON. •
208. Take the equations 2y = 2x + 4,
and 3y = 6x — 12.
From the first we get y = x + 2,
and from the second, y = 2x — 4.
Now, since, by hypothesis^ the y and x in one equation are equal to
the y and x in the other equation, we can equate the values of y, and
make the two it's represent the same thing. Hence, « + 2 = 2a: — 4.
From which we get x =s 6. Solving the first equation with respect to
X, we get aj = y — 2 ; and from the second we get x == ^ + 2.
Now, since the x*s are equal by hypothesis, we have a right to
equate their values. Hence, y — 2 = -^ +2.
Assuming that the y's are equal in the two members of this equa-
tion, we have y =s 8. Hence, the values of x and y are x >= 6, and
y =s 8. It will be seen that these values satisfy both equations.
Elimination by comparison between two equations with two unknown
quantities consists in solving both equations with reference to one un-
known quantity, and equating the values so found. This eliminates
the first unknown quantity, and gives a single equation, from which
the value of the second can be found. The second unknown quantity
being eliminated in like manner, the value of the first can be found.
209. If there had been three unknown quantities, and but two equa-
tions, there would have been, after the elimination of y, a single equa-
tion with two unknown quantities, and the values of these two un-
known quantities would, of course, have been arbitrary. If there had
been two equations and but one unknown quantity, the equation could
not, in general, be satisfied by the same value for that unknown quan-
SQ17ATION8 OF THS TIRST DEGREE. 181
titj. The eqvatioDB a; + 2 » a, and x — 2 « a, cannot be satiBfied
for the same raluee of x.
We gee, then, that the number of equations mtLst he precisely equal to
the number of unknown quantities,
210. The equations have been combined under the Bupposition thai
they were aimidtaneous. If they are not so^ the hypothesis has been
absurd; and the result ought to indicate the absurdity. Take the
manifestly absurd equations;
y = 2x + 2,
and y Bs 2a; — 2.
Combining; we get 2x — 2rr, or Ox = 4. Hence; a; == J =i oo.
The absurdity of the hypothesis is here nointed out by the symbol
of absurdity.
211. If the equations are the same, or differ only in fonU; ihey can,
obTiously; be satisfied by an indeterminate number of values for one of
the unknown quantities; provided that of the other is deduced after the
substitution of the assumed value of the first.
Take; y = 2a; + 2;
and y =8 2x + 2.
Combining; we get Oy =s 0; or y = -jr . Which indicates that y may
have any value whatever. Suppose we assume y ^ 4 ; this value for
y; substituted in either of the equations; will give 2 = 1; and the two
valueS; y SB 4; and x a 1; will satisfy both equations. Assuming
arbitrary values for y, we get an infinite number of values for x, and
the sohition becomes wholly indeterminate.
ELIMINATION BY ADDITION AND SUBTBACTION.
212. Kesume the equations;
2y = 2a; + 4,
and 8y«=6a; — 12.
Multiply the first equation by 3; and the second in like manner by
2; and we will have
6y » 6a; + 12
6y aa 12g — 24
0= 6x — 36.
182 ELIMINATION BXTWEEN SIMULTANEOUS
If these equations are simultaneous, the 6y in the first equation is
equal to the 6y in the second. And, since the x's will also be equal,
the result of the subtraction of the equations, member by member^ will
be = 6:c — 36, or a? = 6, the same value as before found.
Now, to eliminate x in order to find the value of y, the first equation
must be multiplied by 3, and the second must remain as it is.
We will get, 6y = 6x + 12,
and 3^ asB 6x — 12,
and by subtraction, 3^ = 24, ox y = 8.
The same value as when we eliminated by comparison.
It will be seen that the coefficients of the unknown quantity to be
eliminated have been made equal in the two equations ; and, since they
had like signs, the elimination could only be efiected by subtraction.
Had these coefficients, however, been affected with contrary signs, it
would have been necessary to add the equations member by member.
Take as an illustration,
2y — a; = 8
X — y = — 1, by addition, x is eliminated.
F^"2
To eliminate y, multiply the second equation by 2, and we get
2y — X = 3
2x — 2y = -- 2,
and by addition, x = 1
The two values are then ^ = 2, and x :s 1, and these satisfy both
equations.
Elimination by Addition and Subtraction consists in making the
coefficients of the unknown quantity to be eliminated the same in both
equations, and then subtracting the equations member by member, if
these coefficients have like signs, or adding them member by member
if they have unlike signs.
It will be seen that the method of elimination by addition and sub-
traction involves no fractions, whilst the method by comparison, in
general, docs involve fractions. To eliminate by comparison between
the equations
3y = X + 2,
and 2y = X — 2,
will involve the fractions — = — , and — - — •
EQUATIONS or THE FIRST DEGREE. 133
ELIMINATION BY SUBSTITUTION.
213. This consists in finding the value of one of the unknown quan-
tities in one of the equations in terms of the other unknown quantity,
and substituting the value found in the second equation, so as to deter-
mine the value of the second unknown quantity in known terms.
Resume the equations,
2y=:2xH.4y
and 3y = 6x — 12.
From the first we get y = x -f 2 ; and this value of y, substituted
in the second equation (since. the y*a and x's are, by hypothesis, the
same in both equations), gives 3x -f 6 =& 6x — 12. Hence, x =» 6.
In like manner, x, found from the first equation, x =» y — 2, substi-
tuted in the second, gives 3y = 6 (y — 2) — 12, or y = 8. The
two values are then x ss 6, and y = 8, as found by the other two
methods.
This process will also, in general, involve fractions.
ELIMINATION BY THE GREATEST COMMON DIVISOR.
214. This process consists in transposing aU the terms of the two
equations to the first member, and then dividing fhe polynomials in the
first member by each other, as in the method of finding the greatest
common divisor, until a remainder is found which contains but one un-
known quantity. This remainder, placed equal to zero, constitutes the
final equation, and the value of one of the unknown quantities can be
Jeduced from it. The value of the other unknown quantity can be
found by a similar process.
Resume the equations,
2y = 2x 4- 4,
und 3y = 6x — 12,
transposing, 2y — 2x — 4 a 0,
and 8y — 6x-f 12=s0.
Multiplying the second equation by 2, and dividing, we get
2y — 2x — 4
6y — 12x + 24
6y — 6x —12
12
3 Quotient.
— 6x — 36 = 0; or X = 6.
134 ELIMINATION BXTWEEN SIMULTANEOUS
To eliminate x, arraoge witb reference to x, and we have
— 2x + 2y — 4
— 6x + 3y + 12
— 6x + 6y — 12
+ 8 Quotient.
— 3y + 24 = 0, ory = 8
#
It only remains to be shown the reason why the remainder is placed
equal to zero rather than to 10, or anything else.
Let A = represent the first equation after it has been prepared for
division. Let B = represent the second equation. Let Q represent
the quotient after the division of A by B, then,
A |_^
BQ Q
A — QB=:0.
Now, since A is zero, and B zero, A — QB is plainly zero.
For equations of the first degree there will be but one remainder.
But if the equations be of a higher degree than the first, there will be
two or more remainders. But the same course of reasoning will show
that each successive remainder must be zero. For the first remainder
having been shown to equal zero, and the divisor B also equal to zero,
the second remainder must also be equal to zero. Let B' represent the
second remainder, and Q^ the quotient resulting from the division of B
byB'.
Then, B [W^
QW Of
B — Q'B' = 0.
Since B = 0, and B' = 0, plainly B — QB' must be zero.
The third remainder, the fourth, and so on, can, in like manner, be
shown equal to zero.
215. If we combine equations, which are not simultaneous, by either
of the first three methods, the absurdity of the hypothesis is shown by
00, the symbol of absurdity in the result. But when we combine such
equations by the fourth method, the absurdity appears in the final
equation having no unknown quantity.
Take the equations
y = 2x + 2,
and y » 2x + 4.
XQI7ATION8 09 THE FIRST BSaRRE. 185
Combining by laat prooeas^
y_2a; — 2 | y — 2a; — 4
y — 2x — 4
4-2 = 0, wbicb is absurd.
216. Though it is usual to say that the absurdity of combining equa-
tions which are not simultaneous is shown by the final equation, yet
we might retain the trace of one of the unknown quantities, and then
we would still have the symbol oo. In the last example, we might
retain the trace of y, and the final equation would be Oy + 2 ^ ;
whence y = oo.
Remarlcs,
217. Of the four methods of elimination, the last is generally used
when the degree of the second equation is higher than the first, and the
second method (by addition and subtraction) is preferable for simple
equations, since it does not involve fractions. Elimination by substitu-
tion is generally associated with the other three methods after the value
of one unknown quantity is found ; this value is generally substituted
in one of the given equations, and we are thus enabled to deduce that
of the other.
Take the equations,
2y = 2x + 4,
and 3y = fJcc — 12.
From the first we have, y s x -f 2, and from the second ^ as2x
•— 4. Equating the two values of y, we get re + Z = 2x — 4.
Hence, a; = 6. This value for x, substituted in the first equation,
^ves 2y = 16, or y = 8. And substituted in the second, gives
3y = 24, OF y = 8. Hence, the given equations are simultaneous.
But if the substitution of the value for x in the two equations gave
different values for y, we would conclude that the equations were not
simultaneous.
218. Examples in elimination hettoeen tico simple equations of the
first degree involving two unknown quantities.
1. Find the values of x and y in the equations,
and y = a'x + b\
h — y aV — a'h
Ans. X = ;, and y = j- .
a — a a — a'
166 ELIMINATION BETWEEN SIMULTANEOUS
The hypothesis^ a » a', makes both values infinite. The combined
equations show that when a = c^^ a!h must equal aV ; that is, unequal
multiples of the same quantity must be equal, which is absurd. The
equations, in fact, represent two straight lines, and the found values of
X and y represent their point of meeting. The hypothesis, a s=s a',
makes the lines parallel, and their point of meeting is, of course, at an
infinite distance.
Making a ^=i a! and & = &',« and y both become J, or indetermi-
nate. In this case, the two equations become identicsd, and can be
satisfied by any values for x and y, as the symbol % indicates. By this
we mean that arbitrary values may be given to either x or y, and these
arbitrary values for one of the unknown quantities, taken in connection
with the deduced values of the other, will satisfy both equations.
When a = a', and h = Z>', the two lines coincide, and their point
of meeting, being any where on the common line, is^ of course, inde-
terminate.
When h a= 6', we will have x =b 0, and y = 5.
2. Combine 2y + 3x = 4,
and y — 6x = 7.
Ans. X 8= — I, and y cs 3.
3. Combine 2y + 3x = 0,
y — 6a: = 0.
Am. 05 = 0, and y = 0.
When there is no known term or terms in the two equations the
values of x and y will always be zero, since these values will satisfy
both equations.
4. Combine 2y + ~ — 4} = 0.
and 3y-h^^8f.
Am, X S3S 1, and y ss 2.
6. Combine ^ + |. _ l « Q.
and ar -h 2y -f 4 = 0.
Am. X = 32 J, y = — 18^.
6.Combme | + | + f + ^^ = 11.
and 38— |- + |-+y = 36f.
A-M. X ^ 8, and y =s 80.
XQ1TATI0NS or THE VIRBT BBQBBE. 137
7. Combine S^ — 8a; + 6 a 0;
and 7y — 7a? = 14.
Am, y 8 00^ and x ss oo.
8. Combine ^— ^+ 2a! « 0.
and
2 2
o'y 7a'« 4a'
3 3 •
Ans. X ss ^, and y ■= (•
9. Combine — H + 4 = 0.
q 2
and f._^_2«0.
y a;
Ans. X ass -y<^, and y =3 "|.
Regard — ^ or — , as the quantity to be eliminated.
10. Combine 1 4a = 0.
y X
. 3a 2a ^
and = 2a.
y X
Wben one of the unknown quantities is wanting^ it may be written
wiih a zero coefficient.
11. Combine y sb x + 2,
and y = 2, or y s=s Ox + 2.
Ans. X ^Of and y a: 2.
12. Combine x &= 0,
and y == 7x + 4.
Ans. xsszOf and y » 4.
13. Combine a; + y ss a
and ax -\- y ^=h.
. h — a , a' — h
Ans, x = =■, and y «= =,
a — 1 a — 1
What do these values become when a =s 1 ? Why f What, when
h 'Ts af What,, when a = ? What, when h r^ a^f
14. Combine f- ^ = c.
X o
a
and )- y ^s d.
oc — a *^ 1 — 6
12*
188 ELIMINATION BETWEEN filMULT ANEOUS
What do these values become when 5 v 1 ? What^ when c^»df
What, when cssd, and 6 = 1? What, when 5 = 0? How are the
rosolta explained ?
15. Combine 1 es c.
X y
and — H = cr
aV — a'h ^ ah' — a'b
Ans. X = —- T-:, and y = — ; t •
cb' — hd' ^ ad — dc
Solve this example by the four methods of elimination. When x is
eliminated by the last method, the final equation in y ought to be
{ad — a'c) y •\- a'h — ah' = 0.
What do the values of x and y become when a'h = ah' f Why ?
What do these values become when hd ^ cb'f What, when ars^nf^
h = Vj and c= d?
16. Combine y+ -j-y + x = c,
and y — x = c.
. ac . 2be
Ans. x = — TTT and y = rrr— : — -
2h + a ^ 26 + a
The final equation in x, when y is eliminated, ought to be (26 + a)
X •{• oc = 0.
What do these values become when c = ? Why ? What do they
become when a = ? What, when 6 = 0?
17. Combine ^ + y = 2a (1 + a).
X
and y + 2x — 2a =s 2a*.
Ans, X = a, and y = 2a*.
Eliminating by substitution, we get y = 2x'. Hence, the second
equation will give 2x* — 2a* + 2x — 2a = 0, or 2 (x* — a") + 2
(x — a) = 0, or 2 (x — a) (x + a + 1) = 0. Dividing out by the
factors 2 (x + a 4- 1), we have x — a = 0, or x = a. The same re-
sult may be obtained by the fourth method of elimination.
18. Combine ^ + y = 26,
X
and - + 2y «= 26 + a*
X
XQTTATI0N8 OF THB 9IB8T DBaBMS. 189
Combining by fourth method, we have
(y — 2b)x + y I (2y — 26 — a)x-f a.
Preparing for division by multiplying by
(2^ — 26 — g)
«(y — 26)(2y — 26 — a) + y(2y — 26 — a)
x(y — 26) (2y — 26 — a) + ai/ — 2ba
y — 26. Quotient.
2^ — 2ay — 2hy + 26a = 2y (y — a) — 26 (y — a) = 0. Benudnder.
or, (y — a) (2y — 26) = 0, the final equation.
Divide out by 2y — 26, and we have y — a = 0, ory = a; and
this value for y, subetituted for either of the equations, gives x=^ .
We might have divided out by y — a, and then we would have had
2y — 26 =s 0, or y = 6; and this value for y would have given a; = 1.
The equations, then, admit of two systems of values, y ss a, and
X =s ; and y = hj and x ^1. Example 17 gives, also, a
second system, xss — (a + 1), y =s 2 (a + 1)* .
19. Combine
and
X y
y X sa 0.
Ann, a; SB g, and y » g.
20. Combine
and
X y
y-x^2.
Alls.
X
= 2, and y = 4, or a; = — 4, and y = — 2.
Combining by substitution, we get a:" + 2x — 8 as ; or, adding
and subtracting unity, x* + 2a; + 1 — 9 = (a; + 1)« — 3« = (x + 1
+ 3) (a: + 1 — 8) = 0.
By suppressing the first factor, we get x = 2 ,* and by suppressing
the second, we get x = — 4.
21. Combine yx — x = 6,
and X — y = — 2.
An$, X = 2, and y = 4; orx = — 3, and y = — 1.
Id this case, add and subtract }.
140 ELIMINATION BETWEEN 8IHULTANE0T] 8
22. Combine yx = 0.
y
and y + X = 2.
Am, X = 1, and y = 1 ; or x = 3; and y = ^ 1.
AX
23. Combine x + -^ + y — 2,
and a: + 2hy = 0.
' A —46 , 2
-4»w. X = j-T — --5-. , and y = = =-t — —^ .
1 — 6 (a + 2) 1 — 6 (a 4- 2)
What do those yalues become when a = — 2 ? What, when
6 = 0?
24. Combine x + 6 — a + ^ = — ^,
and 2x + — ^ = 2a — 26 + 2.
a — o
Ans, X = a — 6, and y = a — 6-
25. Combine y = ox + 6,
and y = 2.
-Ana. X = , and y = 2.
a
What do these values become when a = ? What, when 6 =i 2 ?
Explain these results.
219. Elimination between any number 0/ simultaneous equations.
The same principles govern the elimination of any number of simul-
taneous equations as have been shown to govern the elimination be-
tween two equations with two unknown quantities. No specific rules
can be given for elimination, because each equation may contain all the
unknown quantities ; or, a part only of them may contain all. It may
be even that no equation, or but one, contains all the unknown quan-
tities. The main thing to be observed is, to eliminate the same un-
known quantity from all the equations that contain it. We will then
have one unknown quantity less than before, and one equation less.
Continue the process of elimination, until a single equation with a single
unknown quantity is obtained. If the number of unknown quantities
is greater than the number of equations it will be impossible to obtain
a single equation with one unknown quantity, because the number of
equations reduced by elimination is always eoual to the number of un-
XQUATIONB OF THE FIRST DEGREE. 141
known quantities reduced. Two eliminations reduce tbe number of
unknown quantities and equations by two ; three eliminations by three,
&c. It is evident, then, that when the number of unknown quantities
exceed the number of equations, the last equation obtained will contain
two or more unknown quantities, and will, consequently, be an inde-
terminate equation.
220. If the number of equations exceed the number of unknown
quantities, it is plain that, before all the equations have been freed
from their unknown quantities, we will get a single equation with but
one unknown quantity. The value of this unknown quantity then can
be determined, and it may not be such as to satisfy all the equations.
221. The number of equations and unknown quantities must not
only be equal to each other, but the equations must be different in cha-
racter , not in forra merely. The equations y=z2x + 2, and 2y = 4x
+ 4, differ only in form, and it is impossible to eliminate between
them.
EXAMPLES.
1. Solve the three equations,
2y — X + « = 2,
2y + 2ac + 4« = 8,
3y + 13x + 3« == 19.
An9, y = 1, X = 1, and « = 1.
r
2. Solve the three eqoations, Y 4 1^1 V *^
Aiu, y=:2fXssS, and « = 4.
S. Solve the three equations,
y -\' X + z = a-\- h-\-c,
y — a: + a + ^ — c ^= a^
y — 2x — a« = a — 26 — 3c.
Am, y =z a, X =&, and z = c.
4. Solve the three equations,
» + 2y = 4,
X + z =s a,
y — z=: b.
Am y = 4 — (a + 6), X ra 2 (a + 6) — 4, and « = 4 — (a + 2&).
142 ELIMINATION BETWBBN BlMXTLTANBOrS
5. Solye the three equations^
X — y = a — h,
X + y + z — c = 4.
Ans. y =^h — 2;SD = a — 2, and « =s 8 + c — (a + ft).
6. Solye the three equations^
2ax + «■ — 2a + -^ +y — a = a — c + 1.
y — 2a; + 3a = a + 3c — 2.
^n«. y s= a, a? = 1, and a = e.
7. Solye the three equations;
' Aa; + By + C« + D = 0,
A'x -f B'y + C'a + ly = 0,
A"x + B"y + C"» + F' = 0.
. _ (A^C^^ — A^^CQ P + ( A^^C — ACKQ 3y + (A(y — AT) IX' '
'• "^ "" (A'B" — A"B') C + (A"B — AB") C + (AB' — A'B) C"
_ (B'^C^ ~ BT^O D + (BC^^ — B"a) jy ■hJi'BV — B(y) D'^
(A'B" — A"BO C + (A"B — AB") C + (AB'— A'B) C"
( A"B^ — A'B") D -t- ( AB" — A"B) JY + (A'B — AB') D^
^ ~" (A'B" — A"B') C + (A"B — AB") C + ( AB' — A'B) CT
What do these become when D, W, and D" are all equal to zero ?
Why ? What do they become when either A, A', A", or B, B', B",
or 0, C, and C" all three are lero ? Why ? What will be the efFect
of making D, JY, and D" zero, and A = A' = A", B = B' = B", and
C = C=:C"?
8. Solye the equations,
X -^ y=z a,
X H- « =« 6.
y + « = <?.
. a -^ c — h a + ft — c h + c^^a
Ans. y = ^ , X = ^ , z = ^ .
What hypothesis will reduce these yalues to zero ? What will make
the first two equal ? What all three equal ?
X =
EQUATIONS OF THE FIRST DEGREE. 14S
9. Solye the three equations^
x-r y + 2z — ^*=4,
x—y — z^2,
2x •\' 2i/ + z t^ a.
Values indetenninate. Why ?
10. Solve the three equations;
a; + y a» rt,
X y
X =* ^>
a o
ftx + cy = 0.
Yalnes found from first and second different from those found from
second and third.
11. Solve the equations^
2y — a — 3 (a — 5) +— = — -^,
c c
y + a; + 2j_4 (a + 5) « — 3 (a + fc),
y — X — 2 (a — h) -f a + z = 2a -\' b.
Arts, y^a-'^h, x=sb — a, and z =s a + h,
12. Solve the equations,
ay 'i' hx ^ c,
y + X'^hy
y + x — 2 ss a.
. c — i* ah — r _ _
Ans. y = -, X = ^ , and » := 6 — a.
a — 6 a — 6
What will be the effect of making & ss a /
13. Combine ll+f + L+f + 11+ ' _ 3.
o 4
X + z x+t z + t
a; — y + < — 2 = 2,
a? + y + l 4-< — 2 = 5.
Ans, y ssl, x^=2, zssS, and ^ a 4.
14. Combine xy = 90,
a; + » = 29,
^n«. X = 9, y = 10, and z = 20.
144 ELIMINATION BETWEEN SIMULTANEOUS
15. Combine |+|. + |+1.«36.
Jn«. a; = 18, y = 20, » = 40, and < = 60.
16. Combine ay =90,
aj + a=29.
a — 1 = 0.
y
Equations absord. Why?
17. Combine x = 2z + 4,
y = 3« — 6,
a: = 2« — 2.
Ans. a; = 00, y = 00, and 2 = oo.
18. Combine a + y + « + < = 4,
a5 + y — ^ — < = 0,
« + t — X — y = 0,
y + « — X — f=:0,
y — sf + a: — <4-to=l.
^r)«. a; = 1, y =1, 2; =1, < s= 1, and w = 1.
19. Combine x + y + » — 4,
2x + 2y + 2« = 5 '
y — « = 0.
j1w«. a; = 00, y = 00, and 2? = oe.
Equations evidently not simultaneous. The combination, then, is
absurd, and the result shows the absurdity,
20. Combine a; + y + 2; r= 3,
X — y — z = 1.
3x + 3y + 32 = 18.
Ans. X = 00, y = 00, and 2: = 00.
The first and third equations plainly conflict Hence, the equations
are not simultaneous.
EQUATIONS OF THX FIKBT DXQBSE. 146
The first and second combined give x = 2; and this value substi-
tuted in the second equation, gives y + 2 =x 1. The values of x and
y + 2, substituted in the third equation, give 9 = 18.
But we may have the absurdity shown by its appropriate symbol by
combining the first and third equations, and retaining the trace of one
of the unknown quantities. Thus, wc may have Ox = 3, or x = oo.
And so, likewise, y and z may be shown to be infinite.
21. Combine the equations,
— X + y -{• z -{- t — h -{• az=(a — 2»)(c-f-rf + c),
X — y — z-\- t = a — h + (h — a) (c + d — c),
ic+ ^ + -t +~=4(a — fe),
c a e
a — b * c (a — b) * d{a — b) e (a — b)
Ans. x = a — b, y =z(a — 6)c, z = (a — b)d, t = (a — b)e.
"When will all the values be zero ? When three ? When the last,
only?
22. Combine the equations,
2y — Sz = 1,
2y + 3z = 7,
y + z = d,
y — 2 = 1.
Ans, y = 2j and 2 = 1.
How does it happen that true solutions are found for four equations,
involving but two unknown quantities ?
23. Combine x + y + z + t + u = 200,
J X V 2 t U
35 + y — 2 = 10,
X + 2 — « = 10,
X -f t — tt = 10.
Ans. x = 20,y = 30,2 = 40,< = 50,andi* = 60.
24. Combine the equations,
« + y + « + < + « = 200,
-4-^ +^ + -4--- 50
x+y — z = 10,
X + 2 — < = 10,
X +/ — tt = 10,
13 K
146 ELIMINATION BETWEEN S.IMULTANEOUS
(A)y + 2; — u = 10,
(B)y + t — u— 20.
Ans. Same as for the last equation.
In the last two examples the new equations, marked (A) and (B),
were not incompatible with the others, and, therefore, the solntions
found, from taking as many equations as unknown quantities, satisfied
the additional equations. When, however, the number of equations
exceed the number of unknown quantities, and the surplus equations
are not satisfied by the same values as the other equations, the solu-
tions will be contradictory.
m
General Remarks.
222. If wc have m equations involving m unknown quantities, we
can find the value of one of these unknown quantities in terms of the
others, and this value substituted in all the other equations will give
us m — 1, new equations, involving m — 1 unknown quantities. We
can now tiiko one of these m — 1 equations, and find the value of a
second unknown quantity in terms of the remaining m — 2 unknown
quantities, and thus get m — 2 new equations, involving m — 2 un-
known quantities. And, by continuing this process," it is plain that,
after m — 1 eliminations, we would get a single equation, involving a
single unknown quantity, and, therefore, would be able to find the
value of that unknown quantity. But, if we have m equations, involv-
ing m + h unknown quantities, we can only make m — 1 substitutions,
and then we will have a single equation, involving h -j-l unknown
quantities. If & = 1, the final equation will contain two unknown
quantities, and bo of the form x + y = 10, an indeterminate equation,
in which the value of x can only be found by attributing arbitrary
values to y. If i = 2, the final equation will contain three unknown
quantities, and be of the form of x -h y + ^ = 10 ; in which x can
only be determined by giving arbitrary values both to y and «. It is
plain, then, that when the number of unknown quantities exceed the
number of equations, the elimination will lead to indetermination.
223. If, on the contrary, we have m -f- 6 equations, involving m
unknown quantities, m of these equations are sufficient to determine
the m unknown quantities. The remaining b equations might, or
might not, be satisfied by the values found from the m equations. If
the h equations are satisfied, they are not independent equations; if they
are not satisfied, they are contradictory equations. Thus, x = 2, and
EQUATIONS OF THE FIBST DEGREE. 147
2x = 4 are satisfied by the same valne of x ; but the second equation
is not an independent equation^ since it differs only in form from the
first. x = 2, and a; = 3 are contradictory equations. We conclude,
then, in general, that the number of independent equations must be
precisely equal to the number of unknown quantities.
224. An artifice sometimes enables us to eliminate between a num-
ber of equations more readily than by the usual direct process. As an
illustration take the equations,
X + y = 5,
as 4- « = 6,
z +!/ = 7.
Let 8=iX + 1/ + z. The three equations then become
8 Z:=i6
«— y = 6
Adding, we get 3« — (x + y + z) = 18. But, since x + y + z
==: 8, we have 2s = 18, or 8 = 9, This, substituted in equations
marked A, gives « = 4, y = 3, and x = 2.
Take, as a second illustration,
xy •■{- Qcz = 27,
y« + y + « = 29,
xyz = 60.
27
By factoring and solving the first equation, we get y + z=z — .
X
27
This, substituted in the second, gives yz ■] = 29. But from the
X
third equation we find yz z= — . Hence, 1 = 29, from which
X XX
27
x = S. Then, y + « = — = 9, and xyz = 60, or vz = 20. Elimi-
X
nating, we get «^ — 9« + 20 = 0, org* — 10» + 25 + « — 6 = 0, or
(z — 5)" + « — 5 = 0, or (2— 5)(« — 5 + l) = 0. Dividing out
Uie second factor, we have loft z — 5 = 0, or « = 5, from which
Take another illnstiaiion,
xy + icz=zl2 (1),
y» + s« = 108 (2),
« — a! = 8 (3).
148 ELIMINATION BETWEEN SIMULTANEOUS
Subtracting (1) from (2), we get i/ (z — x) + z (z — ic) = 96, or
(z — z) (y + 2) = 96 (4). Dividing (4) by (3), member by mem-
ber, we get y + 2 = 12 (5). Equation (2) may be put under the
form of z{y + z) = 108. Dividing this, member by member, by (5),
we will have z = 9. Equation (1), put under the form of a; (y + 2)
= 12, divided by (6), gives x ==1, The value of z substituted in
(5), gives y = 8. Hence, x = 1, y = 3, and z = 9.
PROBLEMS PRODUCING SIMULTANEOUS EQUATIONS OF THE
FIRST DEGREE.
225. Many of the problems already given (Arts. 176 and 206) could
have been solved as readily, or more readily, with two unknown quan-
tities ; and there are many problems which can only be solved by the
use of two or more unknown quantities.
We will solve three of the problems already given, to show the man-
ner of using two unknown quantities.
1. The sum of two numbers is a, and their difference h, what are
the numbers ?
Let X z= greater number, and y = smaller. Then, by the condi-
tion of the problem, x + y = a, and x — y = 6. Eliminating y by
adding the two equations member by member, we^ get jc = ~ -f -j^.
Eliminating x by subtracting the equations member by member, we
have y = — — — . So, we see that when we know the sum and diffe-
rence of two quantities, we get the greater by adding the half difference
to the half sum, and the less by subtracting the half difference from
the half sum.
Thus, the sum of two numbers is 20, and their difference 10 ; the
greater is, by the formula, 15, and the smaller 5.
2. A fox is 125 of his own leaps ahead of a greyhound, and makes
6 leaps to the greyhound's 5; but two leaps of the greyhound are
equivalent to 3 leaps of the fox. How many leaps will the fox take
before he is overtaken by the greyhound ?
Let x = distance passed over by the fox, counted in terms of his
own leaps. Let y = distance passed by the greyhound. Then these
distances would be proportional to the relative number of leaps taken
by the fox and greyhound, if their leaps were equal in length. And
EQUATIONS or THE 7IB8T DEGREE. 149
we would have x, y : : 6 : 5. But since each leap of the greyhound is
equivalent to | leaps of the fox^ the 5 leaps of the greyhound are equi-
valent to 5 . I = ^^ leaps of the fox. Hence, as : ^ : : 6 : \^j and
the distance, y, passed by the greyhound, will be expressed in terms
5x
of the leaps of the fox. From the proportion we get y = -j-, and from
the condition of the problem, y — a; = 125. Combining, we get
X = 500, and y = G25 leaps of the fox.
It will be readily seen that this solution has some advantages over
that with one unknown quantity.
3. Two couriers start from different points on the same road, and
travel in the same direction. The forward courier travels at the rate
of b miles per hour, and the rear courier at the rate of a miles per
hour. They were separated by a distance of m miles at starting.
How long will it be until they come together ?
Let X = distance travelled by forward courier, y = distance tra-
velled by the rear courier. Then, since. the distances traveled jnust be
proportional to the ig^tef of travel, we w ill have x : y : : b : a, or by
= ax. From the conditions of the problem, we have y — x = m;
combining, we cet x = r, and y = -,. These distances, di-
°° a — 6 a — o
vidcd by the rate of travel, will, of course, give the time elapsed before
their junction. Both expressions give for the time, j, as found,
when one unknown quantity was used. This solution, compared with
the preceding, shows how much the use of two unknown quantities
hafi shortened the work.
4. A carpenter wishes to saw a piece of timber, 20 feet long, into
two parts, so that one of them shall be but two-thirds as long as the
other. Where must he place his saw ?
Arts. At 12 feet, or 8 feet from the end.
5. A carpenter wishes to saw a piece of timber m feet long, into two
such parts, that one shall be the — part of the other. Where must he
place the saw ?
Ans, At ; feet, or feet from the end.
6 + o b + a
What values must b and a have to make the solutions in examples
4 and 5 the same ? When will the parts be equal ?
13*
150 ELIMINATION BETWEEN SIMULTANEOUS
6. A colonel wishes to divide his regiment of 800 men, and 10 com-
panieS; in snch a manner that the two flank companies shall contain
each one-third more men than each of the central companies. What
must be the number of men in each flank company, and in each cen-
tral company ?
Arts, 100 men in each flank company, and 75 in each central com-
pany.
7. A colonel wishes to divide his regiment, composed of a men, and
b companies, in snch a manner that the two flank companies shall each
contain — more men than each central company. What must be the
composition of the flank and central companies ?
Arts. Central companies, ea^h ^ ; flank companies, each -^-^ —.
What values must a, b, and c have to make this solution the same
as the last ? What do the expressions become when c = ? What
do the central companies become in that case ?
8. A man becoming insolvent, leaves $4000 to his two creditors : to
one of whom he owes $8000, and to the other $6000. What share
ought each to have out of the $4000 ?
Ans, The first $2285^, the second $1714f
9. A man becoming insolvent, leaves a dollars to his two creditors.
To one he owes b dollars, and to the other c dollars. What share
ou&^ht each to have? . ab . ac
° Ans. X = = — ■ — , and y =r = .
b + t ^ b + c
When will these values becomes equal ? When the first double of
the second? What do they become when he leaves nothing? What,
when he leaves enough to pay his creditors ?
10. A planter hired a negro-man at the rate of $100 per annum, and
his clothing. At the end of 8 months the master of the slave took
him lu:)me, and received $75 in cash, and no clothing. What was the
clothing valued at ? Ans. $12}.
Verify this result.
11. A planter hired a negro-man at the rate of a dollars for c
months^ and was also to give the negro a year's supply of clothing.
At the end of b months the negro was taken away, and the planter
paid m dollars in cash, and gave no clothing. What was the clothing
valued at ? , cm — afe
An$. y = 7 .
EQUATIONS OF THE FIRST DEGREE. 151
When will y = ? When will it be negative ? When infinite ?
The zero solution can be explained most satisfactorily by placing
cm = ah under the form — =-p. • Let the pupil make a problem to
CO
explain the negative solution.
12. A planter has 500 acres of cultivated land, which he wishes to
plant in such a manner that he may have twice as much cotton as com,
and three times as much com as small grain. What division of his
land must he make ?
Ans. 50 acres of small grain ; 150 acres of corn ; and 300 acres of
cotton.
13. A planter has m acres of land and wishes to cultivate it all so
that he may have a times as much cotton as corn, and b times as much
com as small grain. How much of each kind must he have ?
Ans. z = = i r acres of small ffrain, y = ;i i =• acres of
, abm -
com, and x = ^ = r acres oi cotton.
What suppositions upon a, b, and m will make this solution the
same as the last ? What two suppositions will make the three divisions
of land equal ? What will be the effect of increasing b upon the values
of Zf y, and X ? What of increasing or decreasing m f Why does m
enter into the three numerators ?
14. In the year 1692, the people of Massachusetts executed, impri-
soned, or privately persecuted 469 persons, of both sexes, and all ages,
for the alleged crime of witchcraft. Of these, twice as many were pri-
vately persecuted as were imprisoned, and 7|5 times as many more
were imprisoned than were executed. Required the number of suffer-
ers of each kind ?
Ans. 19 executed, 150 imprisoned, and 300 privately persecuted.
15. A planter has $2500 to expend in the purchase of 30 head of
horses and mules. He wishes his horses all to be equal in value, and
his mules all to be equal in value, but each mule to be one-fourth less
valuable than each horse, and the number of mules to be twice as
great as the number of horses. What must be the price of each horse,
and of each mule ? Ans. Each horse $100, each mule $75.
16. A planter has a dollars to expend in the purchase of b head of
horses and mules. He wishes to have c times as manv mides as horses,
152 ELIMINATION BETWEEN SIMULTANEOUS
1
but each mule to be -7 times less valuable than each horse. What
a
must be the price of each mule^ and of each horse ?
Ans. Each horse , ^ , — -. — ^ : each mule — ^ ^ , . ^ e~»
h (d-^ cd — c)' b (a -{- cd — c)
What values must be given to a, b, c, and d to make this solution
the same as the last ? How can the solution be verified ? What will
be the effect of making d = l? Why ? What is the effect of making
c = l. The expressions for the entire cost of the horses and mules
are -= ; , and -,— ^^ — = . What is the effect of making c =
d -^ cd — c' d + cd — c ^
upon these values ? What of making a = ? How must the expres-
sions for the price of each horse and mule be written to show that the
former decreases with the increase of d, and that the latter increases
with the increase of d?
17. The sum of two digits is 6, and the second digit is double the
first. What is the number made up of these two digits ?
Arts. 24.
The digits are the individual figures making up a number. In this
example 2 is the first digits and 4 the second.
18. The sum of two digits is a, and the second digit is b times
greater than the first. What are the digits, and what is the number?
. a ab . lOrt -|- ab
Ans. X = s =, V = -; r : number = — ; — .
b -\- V ^ 6 + 1 6-f-l
The number 24 is made up of the digits 2 and 4, the number 42 of
those digits in reverse order. To express that, 18 added to the first
number would reverse the digits ; we represent by x and y the digits
in the first number. Then, lOx -f- y + 18 = lOy -f x.
19. A number is made up of two digits, and the first is double the
second. If 27 be taken from the number, the digits will be reversed.
What is the number ? Ans. 63.
lOx + y — 27 = lOy + X, and xz=2y,
20. A number is made up of two digits, and the first is a times as
great as the second. And if b be taken from the number, the digits
will be reversed. What is the number, and what are the digits ?
An,. First digit ^^^^ ; second —ZTy Number, ~J^y
(10a + 6) 6
"'' - 9(a-l)
EQUATIONS OF THE VIRBT BEOEEE. 158
What will these results become when a = 1 ? Does the equatioD
of the problem indicate the absurdity '( What will the number become
when bz=a — 1 ? What, when a <[ 1 ? How is the negative solu-
tion explained ?
21. A number is made up of three digits, whose sum is equal to 10.
The first digit is double the second, and the second triple the third.
What is the number ? Ans, 631.
22. A number is made up of three digits, whose sum is equal to a.
The first is b times as great as the second, and the second c times aa
great as the third. What are the digits, and what is the number ?
. a ac ach .
Ans. z =s -, y z= = =-1 X = z r : and
^ lOOacb + lOac -f- a
ri = .
6c 4- r + 1
What will be the effect of increasing c upon the three digits ? What
of increasing b? What upon the digits and number of making
5 = 0?
23. A number is made up of three digits, whose sum is equal to 10.
The first digit is double the second; and if 495 be taken from the
number, the digits will be reversed. What is the number ?
Ans. 631.
24. The sum of the three digits which make up a number is equal
to m. The first digit is a times as great as the second, and if 5 be
taken from the number the digits will be reversed. What are the
digits, and what is the number ?
_ 99am — 6 (a -f 1) __ 99m + b _ (99m + 2>) a
^'*'- ^— (2a + 1)99 ' ^ — (2a + 1) 99' ^ " (2a 4- 1) 99'
, , _ 100a (99771 + ^>) + 10 (99m + ^) -h 99am — b(a-\-l)
and number — ^^^ + 1) 99
25. A gentleman in Richmond expressed a willingness to liberate
his slave, valued at $1000, upon the receipt of that sum from charitable
persons. He received contributions from 24 persons; and of these
there were \^o» fewer from the North than from the South, and the
average donation of the former was |tt« smaller than that of the latter.
What was the entire amount given by the latter ?
Ans, $50 by the former; $950 by the latter.
26. If 7 i be taken from the numerator and denominator of a certain
154 ELIMINATION BSTWEEN BIMVLTANEOUB
fraction, its value will be doubled ; but, if 6 J be taken from tbe nume-
rator and denominator, its value will be trebled. What is the fraction?
Ans. |.
What fraction is that from which, if a be taken from both its terms,
the value will be doubled ; and if h be taken from both its terms, the
value will be trebled ?
. 2a — h ah
A718. z irr, or
4a — 36* 4a — 36
ab
2a — b
The first result gives the reduced fraction, the second (which is
identical with it) the fraction in which the substitution must be made.
We will illustrate by a problem.
27. Find a fraction, such, that if 5 bo taken from both its terms,
the value of the fraction will be doubled ; but, if 4 be taken from both
its terms, the value will be trebled.
Ans. f, or ^
9
The subtraction of 5 from the numerator and denominator of the
second fraction will give |, and the subtraction of 4, in like manner,
will give |. But these results could not be obtained by operating on
the first fraction. The reason of the difference is obvious.
28. If A and C can do a piece of work in 3 days, B and C together
in 7 days, and A and B together in 3|, in what time can each per-
son do the work alone ?
Ans. A in 4^ days ; B in 21 days, and C in 10}.
29. A and C can do a piece of work in a days, B and G can do the
same in b days, and A and B the same in c days. In how many days
can each one, alone, do the work ?
. . . 2ahc . _ . 2a6c _ ^ .
Ans. A in - — ; -^ days, B in —=— r days, C in
bc-^a (b — c) -^ ' ac + b(a — c) ^ '
2abc
bc'^a (6 — c)"
What do these values become when a = cf What, when b=zcf
What, when a=^b = cf Suppose a (b — o) ]> be, will c be a co-
operator^ or a draw-back ?
EQUATIONS OF THE FIB8T DEGREE. 166
80. If A can do a piece of work in 4^ days, B in 21 days, and in
101 dayS; how long will it take them all, working together, to do it?
Ans. a; = 2| days.
Solved by a single equation.
81. If A can do a piece of work in a days, B in 5 days, C in c days,
how long will it take them all, working together, to perform it ?
. ahc _
Ans. a; = -^— days.
ao -^ ac -{- DC
If ft = 0, then, ar = 0, an absurd result. But, by going back to
the equation of the problem, one of the terms is infinite, and the ab-
surdity appears under its appropriate symbol.
32. A farmer has a piece of land, worth $800, and two negroes.
The first negro and land together are worth three times as much as
the second negro, and the second negro and land together are worth
just as much as the first negro. What is the worth of the negroes ?
Ans. First, $1600 ; second, $800.
83. A farmer has a tract of land, worth a dollars, and two slaves.
The first slave and the land together are worth h times as much as the
second slave, and the second slave and land together are just equal in
Talue to the first slave. What is the value of the slaves ?
Ans. First, — ^ :r—^ dollars; second, ^ =■ dollars.
6 — 1 o— 1
What do these values become when 5 = 1?
84. A has a number of five and three-cent pieces in his pocket ; B
wishes to get 24 of them, and gives A one dollar. How many pieces
of each kind must he get ?
Ans. 14 five-cent pieces, and 10 three-cent pieces.
85. A has two kinds of pieces of money in his pocket ; the first
worth a cents each ; and the second h cents each. B wishes to get m
of them, and gives A c cents. How many pieces of each kind must
he get ?
Ans. X = 7- second kind, and y = r- first kind.
a — b ^ a — b
Suppose a = 5, and explain the absurdity of the solution. What la
the meaning of the solution when am = c? What, when am <^ c .
What, when bm ^ cf
36. A has two kinds of money. It takes 8 pieces of the first kind.
156 ELIMINATION BBTWEBN SIMULTANEOUS
and 33i pieces of tlie second kind to be worth a dollar. B offers liim
a dollar for 27 pieces. How many pieces of each kind must he get ?
Ana. 2 of the first; and 25 of the second kind.
37. A has two kinds of money. It takes a pieces of the first kind,
and b pieces of the second kind to make a dollar. A dollar is offered
for c pieces. How many of each kind must be given ?
Arts, —7^ of the first kind, and —7 of the second kind.
— a o — a
Explain the solution when & = a. When h = c: When a = 0,
or c.
38. A certain person has a certain sum of money, which he placed
out at a certain interest. A second person has a less sum by $1666},
which he puts out at one per cent, more interest than the first got, and
receives the same income as the first. A third person has a less capi-
tal than the first by $2857^, but invests it two per cent, more advan-
tageously, and also receives the same income. What are the three
sums at interest, and what the respective rates of interest of each ?
Capitals, $10,000, $8333}, and $7142j|.
Bates of interest, 5 per cent., 6 per cent., and 7 per cent.
39. A gentleman invests a certain capital at a certain rate of inte-
rest. A second gentleman has a less capital by a dollars, but, by in-
vesting it at one per cent, more advantageously he derives as much
income as the first. A third gentleman has a less capital than the first
by h dollars, but, by investing it at two per cent, more advantageously,
he also receives the same income as the first. Bequired the three capi-
tals, and the three rates of interest.
^ . , A aft A 2a (ft — o") ^ ft (ft — a)
An. Capitals, %^^-y %-^-^, »-2a=rr-
lUtes of interest, -^-^^ ^—-^,nnd^^^--^.
Verify these results. Discuss them when ft r= a. When ft = 20,
When ft = 0. When a = 0, &c.
One hundred parts of gunpowder are composed of the following naa-
terials in the following proportions :
For war. For hunting. For mining.
Nitre 75 78 65
Charcoal 12 1 12 15
Sulphur 12} 10 20
100 100 100
EQUATIONS OF THE FIRST DEGREE. 157
40. At the beginning of the Mexican war, the proprietor of the Da-
' pont Mills wished to work up the naaterials of his powder for hunting
and mining, and make war powder out of it. He removed the sulphur
by sublimation, and then wished to ascertain what proportion to take
of the remaining charcoal and nitre in the two specimens. What pro-
portion ought he to have taken ?
Ans. The proportion of the hunting materials to that of the mining,
as III is to z^Q ; or as 125 is to 30 ; or as 25 to 6.
Then, calling 25a; the amount of hunting material in the nitre, we
25 X 75
have 25x + 6x = 75, or aj = |f . There ought then to be — ^ —
6 X 75
parts of the hunting material, and — ^ — parts of the mining material
oJL
to give 75 parts of nitre in the war mixture. A similar relation can be
obtained for the proportion of charcoal.
41. The sum of four numbers is 107. The first, increased by 8, the
second, increased by 4, the third, divided by 2, and the fourth, multi-
plied by 4, will all give equal results. What are the numbers ?
Ans. 20, 24, 56, and 7.
42- The sum of 4 numbers is a. The first, increased by h, the
second, increased by r, the third, divided by d, and the fourth, mul-
tiplied by/, will all give equal results. What are the numbers ?
(a — c)/— (l+/c/)6 (0 + a) /+ (d/+ 1) c — (1 +/d) h
^"'- ' (-2+J)/+l ' (2 + rf)/+l
\2/b + (a — c)/\d 2h-{-a-c
(2 + d)/+l ''^^(2+J)/+r
Verify these results by addition. Show that their sum is equal to a.
Verify them by adding b to the first, c to the second, multiplying the
fourth by/, and dividing the third by d. What single supposition will
make the f rst and second equal to each other ? What single supposi-
tion will make the third and fourth equal ? What will make the first
part zero ? What effect will this hypothesis have upon the other parts ?
43. Four persons owe a certain sum of money: of which the first is
to pay one-third, the second one-fourth, the third one-fifth, and the
fourth one-sixth. After paying a portion of the money, there is still
a deficiency of $36. What portion of it has each to pay ?
Ans. The first, $12il}f ; the second, $9j%\', the third, Z7^^\; the
fourth, $6/;yV
Let 6O2; = the proportion of the first.
14
158 EQUATIONS OF THE FIRST DEGREE.
44. Four persons owe a debt of a dollars : of which the first is to
pay the -^^h part; the second; the — th part; the third, the --7th part;
and the fourth, the — th part. What has each to pay?
c
Arts. The first, —z-. — r— r^ i — -^ r: 5 ^c second,
' ed (c + b) + be (e + d)' '
abde , » . » ahce .
—r-z — —7^ ; — -. r-; thc third, — rr r;^ -, — 7 rrJ the
edic + b)-^ bc{e -{-dy ' ed (c -f ^) + be (c + rf) '
- - ahcd
fourth, —J—. r^- — 7 — - — ^_^.
^ ed{c + b) ^ bc(e + d)
What hypothesis will make the first two results equal ? What the
second two? What all four? What will be the effect of making
either 5, r, d, or e equal to zero. How is this explained ?
45. The denominator of one fraction is 4, and of a second fraction,
8 ; and the numerator of the second fraction is 4 times as great as the
numerator of the first. The two fractions and their greatest common
diyisor, added together, are equal to 3. What are the numbers ?
Am, t, y^, and |.
46. The denominator of one fraction is b, and that of a second frac-
tion is c ; the numerator of the second fraction is m times greater than
that of the first, and the sum of the two fractions is equal to the least
common multiple of their denominators. What are the fractions ?
Am, =■• and
c + w*6' c -h mb'
b c
47. A 1000 cubic inches of bronze were found to weigh 5100 ounces.
A cubic inch of copper weighs b\ ounces, and a cubic inch of tin
weighs 41 ounces. What was the proportion of copper and tin in the
composition of bronze ?
Ans. The copper to the tin as 85 to 15.
48. Some inspectors of cannon weighed m cubic inches of bronie,
and found the weight to be k^ ounces. A cubic inch of copper weighs
b ounces, and a cubic inch of tin weighs c ounces. How much copper,
and how much tin was in the composition ?
.to — mc - , mb — w ^ .
Ans, -= ounces of copper, and -1 ounces of tin.
b — c b — c
What will these values become when b:=.ct Going back to the
VANISHINQ FRACTIONS. 159
/
equation of the problem^ what will 5 = c show in regard to w and mb f
Suppose c = 0, what will the resulta show ? Suppose w = 0, what
will both solutions become ?
VANISHING FRACTIONS.
The symbol J has been interpreted to signify indetermination, and
this is the true interpretation for solutions of equations of the first
degree, when the symbol proceeds from two suppositions, made , either
upon the values found, or upon the equation of the problem. But the
symbol may arise from a single hypothesis, and then it always indicates,
not indetermination, but the existence of a common factor.
Take the expression, —^ which becomes J when=a; — y. But,
, i. . . XV X 1 ^' — y* {^ + y) {^ — y)
by factoring the numerator, we have ^ = ^^-^ — =
•^ . X — y X — y
X '\- y = 2y, when xzE=y, The true value of J in the present in-
stance is'2t/, as shown by removing the common factor, x — y. Again,
-1.1 .a: — y 0, -n . x — y
take the expression —^ ^ = — when x = y. But -5 ^ =
X ^^ 1/ 11
— =0" when x = y. And the true value of
X ~~~ t/
the vanishing fraction is again shown to be finite. But, take
(x — yy
= -pr- when a; = y. Factoring, we get 7 —-^ = = = -jr-.
or 00 when x = y. And the true value of the vanishing fraction in the
rx yY
present instance is infinity. Take again, ^^ ^ = — when x = y.
(x ^"^ 1/ / ix ""~~ V /
Factoring, we get ^^ ^—^ ^ = x — 2/ = when x =z= y. So,
X y
we see, that when there is a common factor existing between the nu-
merator and denominator of a fraction, which factor has become zero,
the fraction may hfeve either of three values, finity, infinity, or zero.
Y (x a)" '
To show it more generally, take the expression ■— =z= — when
X = a. There may be three cases, m may be = w, <[ n, or ^ «.
160 VANISHING PRAOTIONS.
p C^j a)"
In the first case, when w = n, the fraction becomes jr- r- =
P
r^, a finite quantity. In the second case, divided by {x — a)», die
P
fraction becomes /r-7 r — , which is infinite, when x = a. In the
third case, when m"^ n^ dividing by {x — a)", the fraction becomes
— ^^ ^ — = when x = a. And as we have taken a general ex-
pression, we conclude, in general, that a vanishing fraction is one
which assumes the form of {, in consequence of the existence of a
common factor, which has become zero by a particular hypothesis,
and that the true value of the fraction may be either finite, infinite, or
zero.
227. When the common factor is apparent, we have only to strike it
out before making our hypothesis, and we get at once the true value
of the fraction. But there are many expressions, in which it is diffi-
cult to detect the common factor, and it then becomes necessary to
know a process by which the common factor may be discovered. We
will illustrate the process by a simple example, in which the common
r^ ««
factor is apparent. Take the expression, ^ = -r- when a: = a.
Here, the assumed value of x is a. If, however, we make x-=za -\-Ki
and substitute this value for x in both terms of the fraction, reduce the
result to its lowest form, and then make A = 0, it is evident that we
will have done nothing more than attribute to x the value a. Making
the substitution, we get j = -. r= 2a -|- A =
a -f- h — a h
2a when A = 0. The true value of the fraction is then 2a,
ic-^a .
when x=,a.
As the same process is plainly applicable to all fractions in which
the terms are aficcted with numerical exponents, we derive for such
fractions the general
RULE.
Attribute to that term vpon tchich the hypothesis is made tlie value
which reduces the fraction to the form 0/ ^ plus an increment h, reduce
the result to its hwestform and then make h = 0.
VANISHING rOACTIONS. 161
In tlie aboye example, x \a the term upon which the hypothesis is
made; and a the value which reduces the fraction to the form of }
Hence^ by the rule, x = a + h.
228. — EXAMPLES.
Q^ ____ Saa:* -4- Sax* o*
1. Find the value of = -77 when x = a.
X — a
Ans. Sa*.
nS „^^ g/j 4-2
2. Find the value of -5 : — = tt when x = 1.
X* — X — ax + a
Ans.
a—r
8. Find the value of -^^ : = -jr when x = — 1.
x* — ax + X — a
Ans, = — 1^ — .
1 +a
4. Find the value of —5— = -^ when x = m.
XT -}- ax — mx — am U
O + 1?l
c -cwjxi 1 -X' — 4x« — X7/-f 4v" ^
5. Find the value of ^-5 7^ ^ = -tt when x = y.
4x* — 4^x
6. Find the value of -^5 5 — -j- = ~ when x = 1.
XT X* X + 1
Ans. 00.
7. Find the value of the fraction —x -. r- =: -7-- when
x* — bx — ax + ao
X = h. Ans. X = 6 + «•
« Ti. 5 .1^ 1 i. .1^ J. .. ^ — ^3:' — a*x + a*6 .
8. Find the value of the fraction — -. = ;- = -pr- when
X — ox — ax 4- a6
X = a. J.n8. 2a.
14* L
162 VANISHING FRACTIONS.
The two last results arise from the fact of the given fraction being
a double vanishing fraction. It can be put under the form of
^ — ,/ , /, which is a vanishins fraction, when either x = h,
(x — c») (x — a)
or a? = a.
229. It is obvious, in all these examples^ that x, minus the value of
Xf which makes the fraction assume the form of g, is the common fac-
tor. If we, then, divide both terms of the fraction by this common
factor, and then attribute the appropriate value to x, we will have the
true value of the vanishing fraction. But there are many expressions
for which this rule fails, as will be seen more fully hereafter.
Since it is often difficult, if not impossible, without the aid of the
differential calculus^ to ascertain the existence of a common factor, if
there be one, it becomes important to have a simple test by which we
can tell whether ^ indicates in determination or a vanishing fraction.
Take the fraction pry -^~y which becomes J by the single hypo-
thesis x = a. It is evident that the expression is a vanishing fraction,
and that the common factor is some power of x — a. But take the
"P (x a")°
fraction ^-y j—-, which becomes g by the double hypothesis x = a,
and x=zb. It is plain that the expression cannot be a single vanish-
ing fraction like those exhibited in the first six examples ; and if it is
a double vanishing fraction of the form exhibited in Examples 7 and 8,
it cannot be a true solution to a problem of the first degree, since x
cannot have two values. We then conclude, that when g arises from a
single hypothesis upon a solution of an equation of the first degree, it
indicates the existence of a common factor. But, if it arises from
two suppositions, it indicates indetermination. In conformity to this
rule, we have interpreted J, in the problem of the couriers, to indicate
indetermination, because the symbol proceeded from the double hypo-
thesis, m = 0, and a = b.
P0WEB8 AND XZTBAOTION OV BOOTS. 168
FORMATION OF THE POWERS AND
I
EXTRACTION OF ROOTS.
230. The power of a quantity is the result obtained by multiplying
it by itself any number of times.
Any quantity is the first power of itself.
If a quantity be multiplied by itself once, or enters twice as a factor
in the result, the result is called the second power of the quantity.
Thus, 2.2=2*, or 4, and a . a =r a^, are the second powers of 2
and a.
If a quantity be multiplied by itself twice, three times, four times,
&c., or enters into the result as a factor three times, four times, &c.,
the result is ciilled the third power, the fourth power, &c., of the quan-
tity. In general, the number of multiplications of the quantity by
itself is one less than the quantity which designates the power, and the
number of times that the quantity enters as a factor in the result, is
precisely equal to that quantity.
The quantity which designates the power is called the exponent of
the power, and is written a little above and to the right of the given
quantity.
Thus, 2* = 4 is the second power or square of 2.
2' = 8 is the third power or cube of 2.
2* = 16 is the fourth power of 2.
a^r is the y power of a,
and indicates that a has been multiplied by itself y — 1 times, or that
a enters as a factor y times in the expression a^.
When no exponent is written, the first power is always understood.
Thus, 2 = 2», and a + i = (a + i)».
The quantity to be raised to a power may be expressed numerically,
or by letters, and may be entire or fractional, positive or negative.
And^ since the power in every case is a product, we may define the
formation of a power, to consist in finding the product arising from
multiplying the quantity, by itself, a number of times one less then
that indicated by the exponent of the power.
The power differs from an ordinary product y then, in this essential
particular y all the factors of the potoer are equal.
164 FORMATION OF THE POWERS AND
231. The root of a quantity is that quantity which, multiplied bv
itself a certain number of times, will produce the given quantity.
When a quantity, multiplied by itself once, or taken as^a factor
twfce, gives the given quantity; it is called the square root of the
given quantity. Thus, 2 is the square root of 4, because 2.2=4;
and a is the square root of a*, because a . az=za^.
Raising quantities to powers is called Involution.
Extracting the roots of quantities is called Evolution.
Involution deals in equal factors. Evolution finds one of those equal
factors.
232. Involution is a simple process. Evolution is more difficult,
and requires particular explanation. We will begin with the simplest
form of evolution, the extraction of the square root of • whole numbers,
which is nothing more than evolving one of the equal factors out of the
product of two equal factors.
It is evident that evolution is the reverse of involution, and tliat we
cannot extract any ro8t without knowing how the powers of that root
arc formed. To demonstrate the rule, then, for the extraction of the
square root of whole numbers, we must first examine and see bow the
square power of whole numbers is formed.
233. The first ten numbers are
1, 2, 3, 4, 6, 6, 7, 8, 9, 10.
And their squares are
1, 4, 9, 16, 25, 36, 49, 64, 81, 100.
Beciprocally, the numbers of the first line are the square roots of the
corresponding numbers in the second line. We see, also, that the
square of any number below 10 is expressed by not more than two
figures. That is, the square of units cannot give a higher denomina-
tion than tens. So, likewise, it may be shown that the square of tens
cannot give a higher denomination than thousands, since the square of
99 is 9801.
The numbers, 1, 4, 9, 16, &c., and all the other numbers produced
by the multiplication of a number by itself, are called perfect squares.
We see that there are but nine perfect squares between 1 and 100.
The square roots of all numbers lying between 1 and 100 will be found
between the consecutive roots of two perfect squares. Thus, the square
root of 20 lies between the consecutive roots 4 and 5, being greater
than the former, and less than the latter. The square root of 26 lies
between the consecutive roots 5 and 6.
SXTBAOTION 07 BOOTS. 166
The square root of all numbers below 10,000 may be regarded as
made up of tens and units. ThuS; 99, the square root of 9801, is
made up of 9 tens and 9 units. The number 32 is made up of 3 tens
and 2 units. We have seen that the square of a number containing
two figures could not give a higher denomination than thousands ; oon-
Tersely, the square root of thousands cannot give a number containing
more than two figures ; that is, a number containing tens and units.
234. If, then, we have to extract the square root of a number con-
taining more than three figures, and less than five, we know that its
root must contain two figures, and, therefore, be made up of tens and
units. Before we can deduce a rule for the extraction of the root of
thousands, we must know how thousands are derived from squaring
tens and units.
Iiet a represent the tens, and b the units, which enter into the square
root of thousands. Then, (a + by = a* + 2ab + P, Hence, thou-
sands are made up of the square of the tens that enter into its root,
plus the square of the units, plus the double product of the tens by the
units.
Let us square the number 34, made up of 3 tens and 4 units, = 30
-I- 4. Then, 30 corresponds to a in the formula, and b corresponds
to 4.
Hence, a« = (30«) == 900,
2a&=:2. 30. 4 = 240,
ft^ = (4)» = 16,
Then, (34)« == a« + 2ab + 6« = 1156.
235. If we were required to extract the square root of this number,
we would have to reverse the process, and first take out a\ then, 2ah
«Dd&'.
The number, 1156, belonging to the denomination of thousands, its
root must contain tens and units, and
we must first get out the tens by
extracting the square root of the
square of the tens ; that is, the y/ a'.
Now, the square of tens will give
at least three figures, therefore the
root of the tens cannot be sought in
the two right hand figures, and we, 00 Remainder,
therefore, separate them from the
others by a dot, and the given number is then separated into what aie
a« 4- 2ab + i«
11.56
a b
9 00 = a«
30+4
2a + b
2 56 = 2aZ
► + ^
60-t.4
2 40
16
160 FORMATION OF THE POWERS AND
called periods of two figures eacli. The square of 30 is 900; and the
square of 40 is 1600. The number, 1156, falling between 900 and
1600, its root must contain 3 tens plus a certain number of units. We
then extract the greatest square contained in the left hand period, 1100,
and set the root, 30, on the right, after the manner of a quotient in
division. We have now found a, square it, and subtract a* or 900,
from 1156, and the remainder must be 2ab -{-l/: in the present instance
equal to 256. The remainder, 2ah -f 6', can be written (2a + h)b;
and it is evident that, to find b accurately, we must divide by (2a + b).
But, as the teim b within the parenthesis is unknown, we are com-
pelled to use 2a as the approximate divisor. We write 2a, or 60, on
the left as a divisor, and divide 256 by it, and set the quotient 6, or 4,
on the right of the root found, and also on the right of the divisor.
Now, multiply the two terms on the left, 2a + J, or 60 -f 4 by & or 4,
and we evidently form 2a 6 + b', the parts entering into the remainder,
256. Subtracting the two products thus formed from 256, we find no
remainder. Hence, 34 is the exact root of 1156.
We have separated the tens from
the units, to let the begin ner see that
he is really taking a* + 2ab + ft' in
succession, out of the given number,
1156. But we might have indicated
their separation by a point above, and
written 3 4, instead of 30 + 4, and
4 instead of 60 -f 4. When the beginner is familiar with the princi-
ples, he may omit the dots, which are intended to guard him against
confounding the tens with the units. He must observe, however, that
this divisor being the double product of tens, is, itself, tens, and, there-
fore, if written out in full, would contain a cypher on its right. And
since, in dividing by a number whose right hand figure is 0, we point
off that figure from the right of the divisor, and also point off the right
hand figure of the dividend, we must be careful to do this in diyiding
by the double product of the tens. In the present instance, the right
hand figure, 6, of the remainder 256, must be separated from the other
two figures, since, in using 6 as the divisor instead of 60, we have, in
fact, pointed off from the right of the divisor.
236. There is one point of considerable importance that needs some
examination. In getting the second figure of the root, we used 2a as
the approximate divisor of the remainder, 2a& -f- ft' = (2a -f- ft) ft,
a' H- 2aft + ft"
11 . 56
9 00 = a'
a +ft
3*4
2a-f ft
6'4
2 56=2ai
2 bQ = 2al
> + ft»
> + ft'
SXTBAOTION OF BOOTS. 167
whereas, the true divisor, to find h, is plainly 2a + b. Our divisor
heing too small, the quotient, which is the second figure of the root,
can never be too small, but may be too great. It is plain that, when &,
in the expression (2a + b), is very small in comparison with 2a, it
may be neglected. But 6 may be so large that the omission of it
will give too great a quotient. The square of 35 is 1225 ; the square
root of 1225 is then, of course, 35. Now, if we proceed to extract
the square root of 1225, the remainder, after taking out a", or 900,
will be found to be 325 = 2al + h' = 300 + 25. And we see that
the square of the units has added 2 tens to 2ah, the double product of
the tens by the units. When, therefore, we point oflF 5 from the right
of 32 5, and divide 32 by 6, it is plain that the dividend, which ought
to be 2ab (if the divisor is 2a), is too great by 2 tens. The quo-
tient, then, would be too great, if the 2 tens added were divisible by
the divisor. Then the second figure of the root would be augmented
improperly by the quotient, arising from dividing the 2 tens impro-
perly added by 6, or 2a. In the present case, however, if we omit the
2 tens, and divide 30 by 6, we get the same quotient, 5, as when the 2
tens are retained ; their addition has not, then, affected the result.
But, square 19, and we get 361.
In this case, h, in the expression
(2a + 6), is not small in comparison
with 2a, and cannot, therefore, be
neglected without affecting the re-
sult. Now, if we use 2a as the
divisor of the remainder, 2ah -f- 1\
to find 5, we ought to use 2ab alone
as the dividend; and, therefore, if
we use the whole of the expression, 2ab + li^, our dividend is too great.
Dividing 26 by 2, the quotient is 13, which is plainly absurd for the
units of the root. But, we see that 180, or 2aby ought to have been
the dividend corresponding to 20, or 2a, as a divisor. The 26 tens is
then too great by 8 tens ; and, since the 8 tens added, give 4 for a
quotient when divided by the 2 tens of the divisor ; the second figure
of the root is too great by 4, and we must write 9 as that figure, and
not 13. The 8 additional tens in the 26 tens come from the square of
the units, and being divisible by the divisor, 2a, have improperly aug-
mented the second figure of the root. Had the 8 tens not been divisible
by 2a, the second figure of the root would not have been increased at
all, and the quotient of 2a 5 + b* by 2a would have truly been the
a- + 2ah + b*
3 '61
a + b
1 00 — a«
19
2a + 6
26*1 — 2ab + b*
2*9
18 = 2ab
81— 6«
00 Remj
under.
168 rORMATION OF THE POWERS AND
second figure of the root. In general, whenever the square of the units
(&') incorporate into the remainder, 2ab -{' h% tens which are exactly
divisihle hy 2a, the divisor, the second figure of the root will be too
great, and must be diminished by the quotient of the incorporated tens
by the 2a of the divisor.
237. The foregoing course of reasoning has shown that the second
figure of the root may be too great, and the cause of its being too great;
but, since the units of the root are unknown, the number of the tens
proceeding from their square, that are incorporated with 2ah, cannot
be known. We must, then, in practice, form the product of 2a + 6
on the left by the second figure, h, of the root, and compare the result
with the remainder. If the product is greater than the remainder, the
second figure must be diminished until the product is equal to the re-
mainder, or smaller than it. If the given number is an exact square,
its root will be exact, and the product will be exactly equal to the re-
mainder. When the root is not exact, the product must be made less
than the remainder.
The preceding principles enable us to deduce for the extraction of
the square root of whole numbers, embraced between 100 and 10,000,
the following
RULE.
I. Separate the two right hand figures from the other figures or
figure of the given number, and find the greatest square contained in
the left hand period, which may contain but one figure,
II. Set the root of this greatest square on the right, after the mamnet
of a quotient in division. Subtract Hie square of the root thus found
from the first period, and annex the second period to the remainder,
III. Double the root found and place it on the left for a divisor.
Seek how often the divisor is contained in the remainder, exclusive of
the right hand figure, and place the quotient on the right of the root
already found, separated from it by a dot above. Place it also on the
right of the divisor, separated from it in like manner,
IV. Multiply the divisor thus augmented by the second figure of the
root, and subtract the produH from the first remainder. If there is no
remainder, the root is exact. If the product exceed the first remainder,
the second figure of the root must be diminished until the product is
equal to or smaller than the first remainder.
EXTBACTION OF BOOTS. 169
EXAMPLES.
1. Extract the square root of 225. Ans. 15.
2. Extract the square root of 7569. Ans. 87.
8. Extract tlie square root of 2025. Ans. 45.
4. Extract the square root of 841. Ans, 29.
5. Extract the square root of 2500. Ans, 50.
6. Extract the square root of 7921. Ans, 89.
7. Extract the square root ot 9801. Ans, 99.
8. Extract the square root of 4096. Ans, 64.
9. Extract the square root of 5476. Ans, 74
10. Extract the square root of 7056. Ans. 84.
238. We have seen that the second figure of the root has frequently
to be diminished. We may diminish it too much, and it becomes ne-
cessary to know when we have made the second figure too small. The
test of this depends upon the principle, that the difference between two
consecutive squares is equal to twice the smaller number plus unity.
Let a =: smaller number.
Then, a + 1 = consecutive number, or the number just above a.
And (a + 1)» = a« + 2a + 1.
(of = a\
Their difference is 2a + 1, as enunciated.
Now, when there is a remainder, after finding the second figure of
the root, and subtracting the product of it by the quantity on the left
from the first remainder, it is evident that the second remainder ex-
presses the excess of the given number, which we may regard as
(a + 1)*, over the square of the two figures found. If, then, the
second remainder be exactly twice the root found plus unity, it is evi-
dent that the root found is a, and that the second figure of the root can
be increased by unity.
To illustrate, suppose 6 to be the square root of 49, (a + 1)'
then the remainder being equal to twice the root found 49 a
plus unity, the root can be increased by unity. In 36 = a* 6
general, whenever the remainder exceeds twice the 13 = 2a -f- 1
root found plus unity, the root can be augmented by
unity. If the remainder is exactly equal to twice the root found plus
15
170 FORMATION OF THE POWERS AND
unity, tlie root, increased by unity, will be the exact root of the given
number.
239. This role is of importance in finding the square root of imper-
fect squares. Let it be required to find the square root
1 *56 1 12 of 156. We find a remainder 12, and a root 12. Is
1 12 the greatest root contained in 156 ? Is the root 12
2 2 I 56 plus a remainder, or 13 plus a remainder ? By the
44 principle just demonstrated, the true root of 156 must
12 be 12 plus a remainder, because the second remainder
is not double the whole root found, plus unity.
240. This principle also enables us to pass from the square of a
number to' the square of a consecutive number without raising the
second number to the square power. We have only to represent the
smaller number by a, then the consecutive number will be a + 1, and
its square must exceed a* by 2a + 1.
Thus, (lOOy = a« = 10000.
a* (2a + 1)
Then, (101)« = (a+ 1)« = 10000 + 200 + 1 = 10201.
(a + ly a» (2a + 1)
So, also, (102)« = (101 + 1)' = 10201 + 202 + 1 = 10404.
The following are incommensurable numbers.'
EXAMPLES.
1. Find the square root of 1720. An&, 41 + .
2. Find the square root of 14-f5. Ans, 38 + .
3. Find the square root of 6411. Ans, 80 -f .
4. Find the square root of 5555. Ans, 74 + .
6. Find the square root of 1755. Ans. 41 + .
6. Find the square root of 1960. Ans, 44 + .
7. Find the square root of 7777. ' Ans. 88 -f .
8. Find the square root of 6666. Ans. 81 -f .
241. If we square any number, as 12 and 55, containing two figures,
and made up then of tens and units, the square will contain two periods,
counting from the right, and it is plain that the tens can only be
sought in the periods on the left. If we square a number made up of
EXTRACTION OP ROOTS. 171
liandreds, tens, and units^ the square will contain three periods, and
the hundreds can only he found in the left hand period, and the tens
only in the second period, annexed to what is left of the first periods
after the square of the hundreds has been taken from it. In general,
the number of periods in the given number always indicate th^ number
of figure places in the root, and each figure of the root has its appro-
priate period or periods.
The principles that have been demonstrated for the extraction of tho
square root of numbers between 100 and 10,000 can readily be ex-
tended to any numbers whatever. Let a represent the highest denomi-
nation in the root, and z all the succeeding denominations in the root.
Then the number itself will be expressed by (a + %f = a' -f 2as -f a",
an analogous expression to (a -f- l>f = a* + 2a6 -f 2^, and differing
only in its more general significance. The a in one formula is not re-
stricted to represent tens, as it is in the other, but may represent hun-
dreds, thousands, millions, &c. ; and the % of the first formula is not
restricted to represent units only, but may represent tens and units,
hundreds, tens, and units, &c.
Let us square 155 by means of the formula.
Then, a = 100, and s = 55.
a« ==(100y =10000,
2a« = 200'55 =11000,
t^ =(55)» = 3025,
Hence, (155)« = 24025
We see from the formula that a, it matters not what may be its de-
nominatioo, must first be found ; and, that after its square has been
subtracted from the given number, th*e remainder will be 2a« -f s* =
(2a -f «>.
The number, 24025, being greater than lOOOO, its root will be
greater than 100, and, therefore, a*
cannot be found in the two right
hand periods. We seek it in the
period on the left, and after placing
it on the right, and subtracting its
square from the given number, have
14025 for a remainder. We cut off
the right hand figures, because 2, the 000 00 Ecmainder.
approximate divisor, is really 200;
and; after trial, we find 55 to be the right hand figures of the root.
a^ + 2a« + %
240 25
100 00
2
a -f It
155
2a + s
2'55
1 40 25 = 2(
11100 = 2(
30 25 = ««
i«+0«
78
172 FORMATION OF THE POWERS AND
The 55 is set on the right of the root already found, and also on the
right of the divisor. The product of 255 hy 55 is 14025, and there
is, consequently, no second remainder, and the root is exact.
242. The approximate divisor is always large for numhers ahovc
10000, and s can only he found hy repeated trials. But the ahove
process can he greatly simplified hy ohserving that, since 2a enters
into 8, representing several denominations, it must enter into each de-
nomination separately. Thus, in the foregoing example, 2a, or 2,
heing a multiplier of 55, is a multiplier of the first 5, regarded as 5
tens, or 50, and of the second 5, regarded as units. We might, then,
have found the 5 tens and the 5 units, separately taking care to write
the one after the other, so as to make their denomination distinct. In
the present example, 2as being equal to 2 . 55, is, of course, equal to
2 (50 + 5) ; « may he regarded as a single term, 55, to he found at
once, or it may he regarded as made up of 50 and 5, separate terms, to
he found separately. But, if the second and third figure of the root he
found separately as independent numhers, they must he sought for in
their appropriate periods. It will simplify the work when we proceed
in this manner to subtract the square of a from the left hand period,
and bring down each term in succession. In this case, since we make
two terms of «, let « = «' + «". Then the root will be a + «" + «",
and the number will be (a + «" -f s")', which, by performing the mul-
tiplication indicated, will give us a' + 2a^ + «" + 2a«" 4- 2//' + ^
= a« + 2as' 4- j'" + 2 (a + O «" + «"'• When we subtract a« from
the left hand period,
(** + f' + *")* we have, in fact,
a +»'+«" subtracted 10000
^ ^ ^ from the given num-
240 25
1 . . =a«
2a + 8'
2*5
2(a + 0+«"
30 5
14 = 2ay + «'" -f &c. her. After the se-
12 5 == 2a8' + «'* cond period has
1 52 '5 = 2 (a -h /) «" + «"« l>cen annexed to
1 52 5 =r 2 (a + /) s" + «"* t^^ remainder, the
0~" Remainder. 140 truly repreaente
14000, and, sinoe
the zero, on the right of 140, belongs to the denomination of hundreds,
it must be separated from the 14 when we come to seek for the t^is of
the root, because /, the tens, is sought for in 2a8' by using 2a bb the
approximate divisor. Now, 2a/ must be at least thousands, and,
therefore, the denomination of hundreds does not contain a^. We write
«', when found; on the right of the first term of the root and also oo
SXTBAOTION OF BOOTS. 178
tlte right of the diviaor. Multiplying tho divisor thus augmented by
9' in the root^ and subtracting the product from the first remainder, we
will plainly have left, after annexing the next period, 2 (a -f i^y+ /'*.
And we see that the approximate divisor to find /' is 2 (a + <^)- The
whole root already found must then be doubled and used as our approxi^
mate divisor. The right hand figure of 1525 is cut off, because
2 (a -I- «')«" gives at least hundreds. After «" is found, we multiply
2 (a + «') by it, and have no remainder ; the root is then exact. We
have used the broken line, in the above example, after each minuend
and subtrahend to indicate that there were other numbers to follow
them.
(a + ^^ + «" + «"')'
2 98 59 84
1 . . . = a'
a + / + j^' + ^"
17 2 8
2a -h •
27
2(a + + «"
34 2
1 98 = 2a«' + «« + &c.
1 89 = 2aa' + ^^
9 59 =2(a+ «')«" + «'" + &c.
684 =2(a+«'y' + «"*
3448
27 58 4 = 2 (« + «' + «")«"' + «"" + &o.
2 7 58 4 = 2 (ri + 1/ + «")«"' + «"'*.
If we have a number made up of 4 periods, as 2 98 59 84, we know
that its root must contain 4 figure places, which may be represented by
a, a', »", and «'". And the given number must then be equal to
(a -f ^^ + «" + «"')' = a' 4- 2a«' + «^ + 2 (a + «')«"+ «"» + 2 (a + /
We see, then, that 2a is the approximate divisor to find /, the
second figure of the root ; (2a + «') the approximate divisor to find «",
the third figure of the root; and (2(a+«') + «") the approximate di-
visor to find «"', the units of the roots. In other words, we see that
the whole root found has to be doubled to find each figure of the root
succeeding those already found. It is evident, too, that the right hand
figure of each of the successive remainders must be cut off previous to
dividing by 2a, 2 (a + s'), &c., because in all these remainders that
figure is of too low a denomination to make any part of the product of
the figure of the root sought by the approximate divisor. Thus,
8, which belongs to the denomination of tens of thousands, cannot
be a part of the product arising from multiplying 2a, or 2000, by t/,
which is hundreds. This product must give at least hundreds of
thousands.
15*
174 FORMATION OF THE POWERS AND
243. It is plain that, from the manner in which the square of any
number of terms is formed, the foregoing demonstrations for numbers
having two, three, and four periods are general; and we, therefore, have
for the extraction of the square root of any number whatever, the
general
RULE.
I. Separate the given number into periods of two figures each, begin"
ning on the right; the left hand period may contain but one figure,
II. Find the greatest square contained in the left hand period, and
set its root on the right, after the manner of a quotient in division.
Subtract the square of the root found from the left hand period, and to
the remainder annex the second period, and use the number tJius found
as a dividend.
III. Double the root found and place it on the left for a divisor.
Seek how often the divisor is contained in the dividend exclusive of
the right hand figure, and place the quotient on the right of the root
already found, and separate the two figures by a point. Set the quo-
tient also on the right of the divisor and separate in like manner,
IV. Multiply the divisor thus increased by the second figure of the
root, subtract the product from the dividend, and to the reminder
annex the second period of the given number. Use the remainder and
annexed period as a new dividend.
V. Double tlie whole root found for a new divisor, and continue the
operation as before until all the periods are brought down.
Remarks.
244. If the last remainder is zero, the given number is a perfect
square. But, if the remainder is not equal to zero, we have only found
the entire part of the root sought, and the given number is incommen-
surable.
245. If we take 155, the square root of 24025, we observe that 15
has been derived from the two left hand periods. We might, then,
after finding 5, have squared 15, and subtracted its square from 240.
So, after finding the last 5 of the root, we have subtracted the square
of 155 from the three given periods. In general, when we have found
two figures of the root, or three figures, or four figures, &c., we may
subtract their square from the two left hand periods, or the three left
hand periods, &c.
EXTBACTION OF ROOTS. 175
GENERAL EXAMPLES.
1. Extract the square root of 16008001. Atis. 4001.
2. Extract the square root of 4937284. Ans, 2222.
3. Extract the square root of 1111088889. Ans. 33333.
4. Extract the square root of 197530469136. Ans, 444444.
5. Extract the square root of 36000024000004.
Ans. 6000002,
6. Extract the square root of 1259631362889. Ans, 1122333.
7- Extract the square root of 15241383936. Ans. 123456.
8. Extract the square root of 16080910030201.
Ans. 4010101.
9. Extract the square root of 123456787654321.
Ans. 11111111.
10. Extract the square root of 12345678987654321.
Ans. 111111111.
11. Extract the square root of 308641358025. Ans. 555555.
OF INCOMMENSURABLE NUMBERS.
246. An incommensurable number is one whose indicated root can-
not be exactly extracted. Thus^ the y/2^ ^/8, and >/27 are incom-
mensurable numbers. Such numbers are also called irrational num-
bers, and sometimes surds.
"We have indicated (Art. 240,) that the roots of imperfect square
powers were not complete by writing the sign + after the entire parts
of those roots. So we may write the V 5 = 2 + . The number 5
lying between 4 and 9, the square roots of which are 2 and 3, its own
root will be greater than 2, and less than 3. May not this root, then,
be expressed by some fraction whose value is greater than 2, and less
than 3, such as f , or | ? May not the roots of all imperfect square
powers be expressed by vulgar fractions in exact parts of unity ?
To prove that this cannot be, we will demonstrate a theorem upon
which depends the proof of its absurdity.
Theorem.
247. Every number^ P, which will exactly divide the product,
A X B, of two numbers, and which is prime with respect to one of
theni; will divide the other.
176 FORMATION OF THE POWERS AND
Let A be the number with which P is prime. Let Q be the qno
tient arising from dividing AB by P, then -p- =z= Q. We may pat
this equation under the form A X =j = Q. Now, Q is, by hypothesis,
an entire number; the second member being a whole number, the first
member must also be a whole number, else we would have an irreduci-
ble fraction equal to a whole number, which is absurd. The product
of A into ^ has then to be entire'; now, A itself, b entire and prime
with respect to P, hence, p- must also be entire. For a whole number,
multiplied by a fraction, can only give an entire product when the
whole number is divisible by the denominator of the fraction into which
it is multiplied. Thus 4, a whole number, multiplied by the fraction
|, gives an entire product, because 4 is divisible by 2. But 5 into j
does not give an entire product, because 5 is not divisible by 2. Now,
the whole number A is, by hypothesis, not divisible by the denomina-
B B
tor, P, of the expression p-; the product of A by ^ cannot, then, pos-
sibly be equal to the whole number, Q, unless B is divisible by P.
248. We are now prepared to show that the square root of an im-
perfect square, such as 5, cannot be expressed by a fraction. If the
root of 5 can be expressed by a fraction, lot j- be that fraction, a being
greater than h, and prime with respect to it. We assume a and h to
be prime with respect to each other, because, otherwise, their quotient
would be a whole number, and we know that the root of an imperfect
power is found between two whole numbers. We have, then,
>/o = y. From which, by squaring both members, there resolis
a"
5 = -T^. The first member of this equation being a whole number,
o*
the second member must be a whole number also. But -7% cannot be
b
a whole number unless a' is divisible by h ; for, to divide a' by l^, is
to divide it twice by h. Of course, then, if a' is not divisible by b, it
cannot be divisible by &'. But a' is not divisible by />, for, by the fore-
going theorem, a number which exactly divides the product of two
factors, and is prime with respect to one of them, must divide the
EXTRACTION OF ROOTS. 177
oiher. Now the factors of a' are a and a; bia prime with respect
to the first factor, it must then divide the second in order to give an
entirie quotient. This is plainly impossihle, since the second factor is
the same as the first ; a' is, then, not di\isib]e by h ; still less, then,
a'
can a' be divisible by 6'. The equation b z=.-j- must, therefore, be
absurd ] but that equation has been truly derived from the equation
n/5i=—. A correct algebraic operation has led to an absurd result;
but this can only be so when the assumption at the outset, upon which
the operation is based, is absurd. The assumption, y/~5^= j- was ab-
surd, and it has led to an absurd result.
The foregoing reasoning has been in no way dependent upon the
fact that 5 was the particular imperfect power under consideration, and
IB, therefore, general. We conclude, then, that the square root of no
imperfect power can be expressed by an exact fraction.
EXTRACTION OF THE SQUARE ROOT OF FRACTIONS.
249. Since the square of a fraction is formed by squaring the nume-
rator and denominator separately, it follows that the square root of a
fraction must be taken by extracting the square root of the numerator
and denominator separately.
Thus, >/| =1, since | x | =4.
_, / d^ a . a a at
So, also, Kf j-^ = -rf because -7- X -7- = —«•
i»
We may remark that all square roots can be affected with either the
positive or negative sign, if these roots are regarded as algebraic quan-
tities. Thus, the V| may be either -f- 1, or — |, because — | X —
^ = ^. So, also, Ky -r^ may be either + — , or — -=-; since — -r-.
multiplied by itself, gives
78'
EXAMPLES.
1. Extract the square root of -^, Ans. |, or — |.
2. Extract the square root of y^^. Ans, j, or — |.
3. Extract the square root of }. Am. ^, or — ^.
178 FORMATION OF THE POWERS AND
4. Extract the square root of -J. Ans, -J* ®' — v
5. Extract the square root of r^^, Ans. J, or — ^,
6. Extract the square root of ^. Ans, |, or — |.
7. Extract the square root of ^ . Ans. |, or — |.
8. Extract the square root of f J. Ans, j, or — |.
9. Extract the square root of ^^* Ans. y, or — y
10. Extract the square root of y. Ans. 5, or — |
250. These examples show that the positive square roots of proper
fractions are greater than the fractions themselves. Thus, ^Z J ^ J.
The reason of this is plain : the numerator of a proper fraction being
less than the denominator, is not diminished proportionally so much as
the denominator by the extraction of the square root. Let a' and 6* be
two unequal squares, and let a' ^ 6'; then, a ^ 6, when a and 6 are
' both positive. Extracting the square roote of a* and 6', we have
a &nd b for the roots' The extraction of the root has then diminiBhed
a' a fold, and h^ only b fold. The greater quantity has, then, been
diminished the most.
261. The positive roots of improper fractions are less than the frac-
tions themselves, because the extraction of the root diminishes their
numerators more than it diminishes their denominators.
The negative roots of all fractions, whether proper or improper, are,
of course, algebraically less than the fractions themselves.
252. In the foregoing examples the numerators and denominators
were all perfect squares. But, if we have a fraction whose denomi-
nator is not a perfect square, we can readily find its exact root to within
less than unity, divided by the denominator of the fraction. Let it
be required to extract the square root of |. We multiply the nume-
rator and denominator of the fraction by the dcnomfnator, which does
not alter the value of the fraction, and we have s/'i = \ / = =■ =
• V 5 X 5
v'^J = 5 +. l*he root of the numerator lies between 6 and 7. The
root of the fraction is greater than j{, and less than { ; and we see that
I is the true value of the fraction to within less than ^. By this, we
mean that when we take f as the true root of the fraction, we commit
an error loss than J. We can, of course, get a nearer approximation to
the true value of the fraction by multiplying both terms of the fraction
by the third, fifth, seventh, or some odd power of the denominator. This
will make the denominator a perfect power, and its root can be exactly
XZTRAGTIOK OF BOOTS. 179
* A mv ,-5- /8x5x5x5 /lOOO 81 . _
found. Thus, VT= \/ 5^5^5^5 = \/-625 = 25 +" ^«
true root is greater than |^, and less than ||. Hence^ |^ differs from
the true root by a quantity less than ^^.
253. When the denominator and numerator are both imperfect
powers, as in the example just given, we may make the numerator a
perfect power by multiplying both terms of the fraction by the first, or
some odd power, of the numerator. But, in this case, the degree of ap-
prozimation is not immediately apparent. Thus, >/^= \/ "K — & ^^
I i^prozimatively. The true root is greater than |, and less than |.
The degree of approximation can only be determined by reducing these
fractions to a common denominator. We have, then, ||, and |§, and
iheir difference is ^. Then, || differs from the true root by a
quantity less than :^\. And ||, or g, also differs from the true root
by a quantity less than ^^. It is plain that the degree of approxima-
tion can be more readily determined by making the denominator ra-
tional than by making the numerator rational. It is eyen preferable to
make the denominator rational when the numerator is already so,
though the process of making the denominator a perfect square make
the numerator irrational. Thus, to find the approximate root of f , place
264. If the denominator is already rational, we have only to extract
its root for a new denominator, and write over it the approximate root
of the numerator for a new numerator.
We have, then, for finding the approximate root of any firaotion^ both
terms of which axe not rational, the following
RULE.
Make the denominator rcUtonaly if not already so, hy multiplying
both terms o/^ fraction by the first, or some oddpotcer of the denomi-
natoTf according to the degree of approximation required, so that the
denominator of the given fraction shall be the square poioer of the de-
nominator of the fraction that marks the degree of approximation.
Then extract the root of the denominator for a new denominator, and
write over it the approximate root of the numerator for a new numera-
tor. Affect the new fraction with the double sign, to indicate that there
18Q FORMATION OF THE POWKES AND
are ttco roots equal, tcith contrary signs. If the denominator he
already rational^ and a greater degree of approximation is required
than that indicated by unity divided by the root of the denominator,
multiply both terms of the fraction by the third^ fifth, seventh poicer,
dec, of the denominator, according to the degree of approximation
required, and proceed as before.
EXAMPLES.
1. Extract the square root of } to within less than } of its true
value. Ans. db }.
2. Extract the square root of } to within less than \ of its true
value. Ans. ± %.
8. Extract the square root of f to within less than \ of its true
value. Ans. ± }.
4. Extract the square root of | to within less than y*^ of its true
value. Ans. dz ||.
5. Extract the square root of | to within less than \ of its true
value. Ans. d: |, nearer ^ than |.
6. Extract the square root of | to within less than ^-^ of its true
value. Ans. d= ^f •
7. Extract the square root of ^ to within less than ^^j of its true
value. Ans. =b ^^.
8. Extract the square root of -^ij to within less than yj^-^ of its true
value. Ans. ±: ffj.
In the first example^ the multiplier of both terms of the fraction is
2 ; in the second, (2)* ; in the third, unity ; in the fourth^ (4)' ; in
the eighth, (27)».
255. Since the denominator of the fraction may be raised to as high
Hu even power as we please, it is evident that the degree of approxi-
mation can be made as close as we choose to the true value of ihe
fraction.
256. We may, by a similar process, determine approximatively the
roots of incommensurable numbers to within less than unity, divided
by any whole number.
Let it be required to determine the square root of 2 to within lees
EXTRACTION OF BOOTS. 181
than J of its true value. Then, VT= \/ r6V ~ "^ = I +•
The true value lies between § and |y and, therefore, | differs from the
true value by a quantity less than J. We multiplied and divided the
given number by the square of the denominator of the fraction that
marked the degree of approximation required. This, of course, did not
alter the value of the given number, it simply placed it under the form
of a fraction with a rational denominator. The next step was to ex-
tract the root of the numerator to within the nearest unit, and to write
the result over the exact root of the denominator.
To demonstrate a general rule applicable to any number, and true for
any degree of approximation, let a be the number, and — the fraction
n
that marks the degree of approximation. Then, >/a = \/ g — .
Let r denote the root of an^ to within less than unity ; in other
words, let r denote the entire part of the root of an^. We will then
/on"* /T«" r ,^ /(r + 1)" r + 1
have ^a = ^/ ^ > V ^' ^' n' ^^^ < V ^?~' ^' "^^
T r+ 1
The true root of a, then, lies between the numbers — - and , which
n n
1 T
differ from each other by — . Hence, — differs from the true root by
'' n n
1 r-f 1
a quantity less than — . So, also, differs from the true root by
1 . . r
a quantity less than — . We then have a right to take either — , or
r+1
— ■ — , for the approximate root. That one is taken to which the root
lies nearest.
To find the approximate root of any number, a, to within less than
— of its true value, we have the following
»
RULE.
Multiply and divide the given number by the square of the denomi-
nator of the fraction that marks the degree of approximation. Ex"
trtict the root of the numerator of the fraction thus formed to within the
nearest unit, and set the result over the exact root of the denominator.
Give the double sign to the root.
16
182 FORMATION OF THE POWBRS AND
EXAMPLES.
1. Extract the square root of 2 to within less than ^ of its true
value. Afis. d= J.
2. Extract the square root of 50 to within less than i of its true
value. Ans. d= ^®.
3. Extract the square root of 50 to within less than ^^^ of its true
value. Ans. ± ^-^^f,
4. Extract the square root of 50 to within less than y ^^ of its trde
value. Ans. ± ^J.
It is ohvious that, hj increasing the denominator of the fraction that
marks the degree of approximation, we may make the approximate in-
definitely near to the true value of the root of the given numher.
257. Approximate roots of whole numbers expressed decimally.
To extract the root of any whole number, a, to within any decimal
of its true value, we have only to change the decimal into' an equiva-
lent vulgar fraction^ y^, yj^, or whatever it may be, and then multiply
both terms of the fraction by 10", 100', &o. Whatever the decimal,
which marks the degree of approximation, may be, it can be changed
into a vulgar fraction, ■ ] in which m is a positive whole number,
greater by unity than the number of cyphers between the decimal point
and first significant figure. Thus, -1 = TTcyTv ^°^ w s= 1 ; -01 « -PtT^
and m = 2 ; '001 = ttttts, and m = 3.
Hence, Va = \/ —
(10)
2i
Representing the entire part of the root of a (lO)*" by r, there re-
r + l
^"^^ \/wy? > mv-' ^°^ <
(lOy- ^ (10)-' ^ (10)-'
r 1
Hence, ^ .^ is the true root to within less than . ; that ia,
r 1
differs from the true root by a quantity less than -^jr — . Now,
(10)- ^ ■> ^ - • - (10)--
multiplying the given quantity, a, by (10)'", is the same as annexing
XXTEACTION Of E00T8. 188
2m cyphers } for^ multiplying it by (\0y, annexes 2 cyphers ; by (10/,
annexes 4 cyphers ; by (10)*^ 8 cyphers, &c. And dividing the ap-
proximate root found, r, by (10)"* is plainly the same as cutting off
from the right of the root found m places for decimals, for to divide
the root by (10)', (10)*, (10)*, is the same as cutting off from the right
one, two or three places of decimals. Hence, to approximate to the
true root of any given number to within a certain number of decimals,
we have this
RULE.
Annex twice a$ man^ cyphers to the fftven number as there are deci-
mal places required in the root, extract the root of the number thus in-
creased to within the nearest unit, and ctU off from the right the re-
guired number of decimal places.
EXAMPLES.
1. Eequired >/2 to within -1. Ans, 1-4.
2. Required >/2 to within -01. An^, 1-41.
3. Required v/'2 to within 001. Ans. 1-414.
4. Required >/b5 to within 01. Ans, 7*07.
5. Required >/"50 to withili -001. Ans, 7071.
6. Required v/50 to within 0001. Ans, 70710.
7. Required x/9000to within -1. Ans, 94-8.
8. Required x/ 9000" to within 01. Ans, 94-86.
9. Required >/ 9000" to within 001. Ans, 94-869.
10. Required vTS5"to within -01. Ans, 12-04.
11. Required s/T46"to within -001. Ans. 12041.
12. Required VlOOO to within -001. Ans. 31-622.
18. Required V"lOOO to within 0001 Ans, 31-6227.
14. Required >/ 100000 to within -01. Ans. 316-22.
15. Required y/ 100000 to within .001. Ans, 316-227.
Examples 12 and 14 show that, to pass from the root of any number
to the root of a number 100 times as great, we have only to remove the
decimal point one place further to the right. The converse is evi-
dently true also.
184 FORMATION OV THE P0WER8 AND
MIXED NUMBERS.
258. The approximate root of mixed numbers can now readily be
found. Suppose it be required to find the approximate root of 2-5 to
within -1. If we annexed two cyphers, as before, the result would be
2500 ; and then, when wo shall have come to point ofif into periods, the
whole number, 2, will be united with the decimal 5. The root found will
be 5, which is plainly absurd. But, 2*5, changed into an equivalent vul-
gar fraction, is f ^. Hence, by the rule for vulgar fractions, >/2'd =
y/^^ = VfgJ = ig-|-z=l-5 + . We see, that in the present instance,
we have annexed a swingle cypher, which made the decimal places even,
and double the number of places required in the root. We next
pointed off from the root the number of decimal places required. If we
are recjuired to find the approximate root of 2-5 to within j^^, y J^, or
•01. Then, >/2-^ = ,/l8 = >yfHA^ = ig§ = 1-58. We
have, obviously, added three cyphers to 5, and, therefore, made the
number of decimal places even and equal to the number of places re-
quired in the root. In pointing off for decimals, we have only pointed
off two places, the number required in the root. To demonstrate
the rule in a general manner, let a be the entire part of the mixed
number, and ^^ the decimal part. Then the given number will be
ah
(r r-T— =
10™ ~" lO"- '
^ / b /ah /abx (10)-
the entire part of the root of a X 6 (10)". Now, the multiplication
of h by (10)"" is the same as annexing m cyphers to ft, and whenever
m is odd, the number of decimal places will be even, and double the
number required in the root. When m is even, the decimals will not
be mixed with the whole numbers, as in the mixed number 3*45| and
there need be no cyphers annexed.
We have supposed, in the general demonstration, that the number
of decimal places required in the root was precisely equal to the num-
ber of decimal places in the mixed number. But, if this were not the
case, the denominator of the equivalent vulgar fraction has only to be
multiplied by such a power of 10 as will make it 10 with an exponent
twice as great as the number of places required in the root. The na-
EXTRACTION OF BOOTS. 185
merator^ when multiplied by this power of 10^ will have its number of
decimal places even; and equal to double the number of places required
in the root.
RULE.
Annex cyphers until the number of decimal places in the mixed
number is even, and equal to double the number of places required in
the root. Extra>ct the root of the result to within the nearest unit, and
then point off from the right, for decimals, the number of decimal
placet required in the root.
EXAMPLES.
1. Extract the square root of 4*9 to within •! Am. d= 2*2.
2. Extract the square root of 4-9 to within '01. Ans. ± 2*21.
3. Extract the square root of 4*9 to within '001.
Ans. ±2-218.
4. Extract the square root of 4*25 to within -01.
Ans. ±206.
5. Extract the square root of 4*25 to within -001.
Ans. ±2061.
6. Extract the square root of 96*1 to within -1. Ans. ± 9*8.
7. Extract the square root of 9*61 to within '1.
Ans. ±31 exactly.
8. Extract the square root of 146*755 to within 01.
Ans. ± 1207.
9. Extract the square root of 14575-5 to within 4.
Ans. ± 120-7.
10. Extract the square root of 101-7 to within -01.
Ans. ± 1008.
11. Extract the square root of 1001-01 to within -1.
Ans. ±31-6.
12. Extract the square root of 10*0101 to within -01.
Ans. ± 3-16.
13. Extract the square root of 1728-555 to within 01.
Ans. ± 41-57.
14. Extract the square root of 172855-5 to within -1.
Ans. ±415*7.
16*
186 FORMATION OF THE POWERS AND
15. Extract the square root of 17285550-666 to within -01.
Am. dr 4157-58.
16. Extract the square root of 1728555066-6 to within 1.
Am, ±41575-8.
ROOTS OF NUMBERS ENTIRELY DECIMAL.
■
259. Let it be required to extract the square root of -4. This deci-
mal, changed into an equivalent vulgar fraction, is ^V* Hence, ^^ -4 =
y/rfy^ = VtV^ > A> *°^ <1 T^« Hence, -6 is the approximate root
to within less than -1.
We see that the number of decimal places was made even before the
root was extracted.
If we were required to extract the square root of -4 to within -01,
•001, the denominator of the equivalent vulgar fraction must be made
(loy, (10)'.
The decimal fraction can be written =^.
r + l
_ / b Ihx (10)" . r , ^
H«°<«' V 10^ = V -aor" > (10)=' "•* < (lor •
It is plain that the multiplication of h by (10)" makes the number
of decimal places even. In pointing off for decimals, as many places
must be cut off from the right as there are periods.
RULE.
Annex cyphers to the given decimal until its places are even. Ex-
tract the root of the result, as in whole numbers, and cut off from the
right, /or decimals, as mani/ places as there are periods in the numUr
whose root was extracted. If it he required to extract the root
within a certain decimal, annex etchers until the number of periods is
equal to the number ofpluces required in the root,
EXAMPLES.
1. Required \/ -25. Ans. db -5 exactly.
2. Required >/ -9 to within -1. Ans, ± *9.
3. Required >/ -09. An$, dr *3 exactly.
4. Required v^-009 to within -01. Ans, =b -09.
EXTRAOTIOir OP ROOTS. 187
6. Reqnirod V'^OOOQ. Arts. =fc -03 exactly.
6. Required ^ -725. Ans. =t *85 exactly.
7. Required Vl)725 to within -01. An$. =fc -27.
8. Required V 00725 to within -001. u4n«. =fc -085.
9. Required >/"^. jln«. db -9 exactly.
LO. Required V0081. ^ Am. =h 09 exactly.
11. Required v^ -00081 to within -001. Ana. d: -028.
12. Required ^ 000081. Am. ± -009 exactly.
13. Required y/ 0000081 to within -0001. Ans. db 0028.
L4. Required %/ '0144. Ans. ± *12 exactly.
15. Required >/ 04937284. Ana. dt -2222 exactly.
L6. Required y/ -05555555 to within .0001. Ans. ± -2857.
17. Required V -1111088889. Ans. db -33333 exactly.
These examples show that the roots of decimals are greater than the
decimals themselves. This ought to be so, for all purely decimal num-
bers can be changed into proper fractions.
SQUARE ROOT OF FRACTIONS EXPRESSED DECIMALLY.
«
260. Vulgar fractions may be changed into decimal fractions, and
then their roots may be extracted by the last rule. If the given frac-
tion be mixed, it must be first reduced to an improper fraction and
then changed into an equivalent decimal fraction.
EXAMPLES.
1. Required V| = n/-6666 =+-81. Ans. d= -81.
2. Required >/ i = i/^. Ans. d= -5 exactly.
3. Required V^ = V 11111111. A7is. =k -3333.
4. Required V^b = >/ '0625. Ans. =fc -25 exactly.
5. Required Vjj = V 012345679012. Ans. ± -111111.
6. Required \/ i to within -01. Ans. =t -86.
7. Required y/'^l to within 01. Ans. =fc 1-58.
8. Required V2i. Ans. db 1*6 exactly.
9. Required v^TT ^ ^^^^^ '^^' ^"*- =^ '^^^'
188 FORMATION OF THE P0WEB8 AND
10. Required -J 12^ to within 001. Am. ± 8-523.
11. Required v^25^to within 001. Ans. ± 5-011.
12. Required >/b^ to within -01. An%. db 2-25 exactly.
The forcgoiDg examples show that a vulgar fraction, which is a perfect
square, may or may not have an exact root when changed into a deci-
mal fraction.
261. — GENERAL EXAMPLES.
1. Required V48303584-4856 to within -001.
Am. dr 6950-078.
2. Required v^25012001-44 to within 1.
Ans. d= 5001*2 exactly.
3. Required >/ -0289 to within -01. Am. d= -17 exactly.
4. Required V 000144 to within -001. Am. db -012 exactly.
5. Required y/^ to within -001. Am. ± -242.
6. Required \/^.to within -OOl. Am. =b 425 exactly.
7. Required V"^ to within -0001. Am. db -0625 exactly.
8. Required y/ y^'j^ to within -00001.
Am. =t -03125 exactly.
9. Required V"T728 to within -001. Am. d: 41-569.
10. Required x/ 1728 to within tJ,. Am. db ^Jff.
11. Required >/l6|I| to within -001. Am. ± 4103.
12. Required y/-^-^ to within -1. Am. ± -2 exactly.
13. Required yf^ to within -00001. Am. db -03703.
14. Required sf^i to within -00001. Am. d= -01234.
EXTRACTION OP THE CUBE ROOT OP NUMBERS.
262. The cube or third power of a quantity is the product arisbg
from taking the quantity three times as a factor. Thus, the cube of
mismXmxms: m'. The cube root of a quantity is one of the
three equal factors into which the quantity can be resolved. The pro-
cess of extracting the cube root consists, then, in seeking one of the
equal factors which make up the given quantity. When the quantity
can be exactly resolved into its three equal &ctors^ it is said to be a
SXTRAOTION OF ROOTS. 189
perfect cube. But, wlien one of its factors can only be found approxi-
matiyelj; it is said to be an incommensurable quantity, or an imperfect
cube. Thus, ^/S, and ^27 are perfect cubes; but v^9, and v^ 26 are
incommensurable^ or imperfect cubes.
The first ten numbers are
1, 2, 3, 4, 5, 6, 7, 8, 9, 10.
and their cubes
1, 8, 27, 64, 125, 216, 343, 512, 729, 1000.
Reciprocally, the numbers of the first line are the cube roots of the
numbers of the second line.
We jsee, by inspection, that there are but nine perfect cubes among
all the numbers expressed by one, two, and three figures. All other
numbers, except the nine written above, expressed by one, two, or
three figures will be incommensurable, and their roots will be expressed
by whole numbers plus irrational parts, which can only be determined
approximatively. Thus, the ^9 consists of the whole number 2, plus
an irrational number. Because 9 lies between 8, whose cube root is 2,
and 27, whose cube root is 3. By a course of reasoning similar to that
already employed (Art. 248), it can be shown that the cube root of an
imperfect cube, as 9, cannot be expressed by an exact vulgar fraction.
For, if it can, let -j- be that vulgar fraction ; then, yW= -=-, or 9 =
n-. But, if -=- be an irreducible fraction, from what has been shown,
tr b
— must be an irreducible fraction, and we then have a whole number
equal to an irreducible fraction, which is absurd. And, since the
generality of the reasoning has not been affected at all by the selection
of the particular number, 9, we conclude that the root of an imperfect
cube cannot be determined exactly.
Before demonstrating a rule by which the roots of perfect cubes can
be determined, or the roots of imperfect cubes found approximatively,
it will be necessary to examine the manner in which a cube, or third
power, is formed. When the number contains less than four figures,
its cube root, or the entire part of that root, must be found among the
first nine figures. When the number contains more than three figures,
its rt)ot must be made up of a certain number of tens and units. Let
a = tens of the root, and b = the units of the root. Then, the number
will be (a + ly. And, by actually multiplying (a + h) the required
190 FORMATION OF THE POWERS AND
nnmber of times, we will get (a -f ft)* = a* + Mb + Soft* + 6^.
That utj a number, whote root is tnade up of tern and uniti, is equal to
the cube of the tens, plus three ttmei the square o/ihe tens by (he unitSy
plus three times the product of the tens by the square of the units, pf us
the cube of the units.
The formula may be written (a + by = a* + (Sa* + 3a6 + I/)b.
And we see that the first thing to be done is to extract the cube root
of a', then the tens (represented by a) will be known. It is obvions,
too, that the true divisor of the remainder, after a' has been taken out,
to find b, or the units, is the coefficient of b, (3a' + 3a5 + ^) > ^u^
since b is unknown, the last two terms of thia coefficient, Sab and I/,
are unknown. Hence, we are compelled to use 3a' ajs an approximate
divisor of the first remainder to find b. Now, b, thus found, will, in
every case, be too great, because the divisor has been too small ; bat it
may happen that when b is small, the addition of 3a6 + b^to 8a' would
not diminish the quotient b by unity, and then there will be no error
committed in assuming 6 to be the true quotient^ provided we increase
our divisor by 3a6 + &', and form the parts that enter into the fiist
remainder.
We will illustrate by an example. Let it be required to extract the
cube root of 1728.
We begin by separating the three right hand figures, because the
a + 6 cube of
3a' =300 1'728
(3a+5)6=:X30+2)2= 64 1000 =
(3a« + 3a6+£') =364 728 = (3a«+3a5+y)i ** ^^^
728 = (3aH3a6+&')6 ^^^^usands,
and,tibere-
fore, cannot be contained i^ the right hand period. We next seek the
greatest cube in the left hand period, which is really 1000; the root is
one ten, or 10, we set it on the right, after the manner of a quotient in
division. 3a', or 300, is assumed as an approximate divisor of the
remainder after a', or 1000, has been taken from 1728. The remain-
der, 728, being represented by (3a' + 3a^ + b^b, the true divisor to
find b is, of course, the parenthetical coefficient of b. Having found h,
or 2, by means of the approximate divisor, we set it on the right of the
ten, separating it by a point to indicate that it is of a different denomi'
nation. We next add (3a -f b)b, or 64, to 3a', or 300, and we have
864 for the true divisw, provided that b has been fonnd oorrectly.
This divisor we will oall the supposed true divisor. Finally, we multi-
1*2 the tens
a' must be,
SXTEAOTION OV ROOTS. 191
plj (3a' + Bah + h*, the wppottd true divuor by b, and the prodnct
xoade np (Sa' + 3a5 + h^)b, the parts entering into the remainder.
The product thns formed being exactly equal to the remainder, proved
two things : first, that if the tme, instead of the approximate, divisor
had been nsed, b wonld not have been diminished by unity; and,
second, that the number 1728 is an exact cube, and that its root is
12. In the present example, it is easy to see why b was found
correctly by using the approximate instead of the true divisor; a and
b being both small, the omission of Sab + I^ did not mat-erially affect
the divisor.
We have, from the foregoing, a simple test by which to ascertain
whether &, found by using the approximate divisor, 3a', must be di-
minished. Whenever the supposed true divisor will give a less quo-
tient than bf we conclude that b was too^ great, and it must be dimi-
nished by 1, 2, &c., until the supposed true divisor will enter into the
remainder the same number of times as the approximate. It will never
happen when the unit figure of the root is small with respect to the
tens, as in 71, 82, 93, &c., that the b, found by using 3a' as a divisor,
must be diminished by unity, or some greater number. But, when the
unit figure is great with respect to the tens, as in 18, 19, &c., it may
be necessary to diminish b by one or more units. The reason of this
is plain. We will illustrate by an example. Kequired the cube root
of 5882.
We see that b, found by using 3a' as a divisor, is 16, and the sup*
posed true divisor will
then be 1036 ; but this, 5832
instead of entering 16 3a« = 300 1000
times bto the remainder, (80 + 16)16 = 736 4832
4832, wiU only enter 4 (1036)16 = 16576
times. Besides, 1036, or
the supposed (3a' + Sab + ft^, when multiplied by b, will give a pro-
duct, 16576, greater than the remainder. The unit figure b then too
great, and has to be diminished by 8 ; this diminution can only be
determined by trial. The process ought to have been
5832 I a + 5
1000 1 1"8
3a' = 300 4882 = C3a' + 3a6 + b^b
(30+ 8)8 = (3a + 5)5 = 304 4832 = (3a' + 3a6 + 6^6
3a' 4- Sab + 6' = 604
192 FORMATION OF THE POWERS AND
The true divisor by trial was found to be 604. This divisor was repre-
sented by (3a' + Sab -h ft*), and, when mnltiplied by b, or 8, gave
(3a' H- 3a6 + b^by the remainder.
Take, as another example, 6859.
6 859
3a« « 300 1 000
a+ft
19
(3a + 5)6=(80 + 9)9 = 361 5 859 =
3a« + 8a6 + 5« =. 651 5 859 = |
(3a« + Sab + b'yb
[Sa^ + Sab + ft")6
By trial, 9 was found to be the second figure of the root, the sup-
posed true divisor was then determined to be 651, and this, multiplied
by b or 9, gave 5859. The given number was then a perfect cube, and
19 its exact root. In this case, the supposed true divisor is actually
the true divisor.
263. The process for extracting the cube root of a number below
10000 may be without any difficulty extended to all numbers whatever.
Suppose the number to be 1881365963625, its root may still be re-
garded as made up of tens and units, the cube of the tens cannot enter
into the last three figures, 625, on the right, and they may, therefore,
be separated from the other figures. The greatest cube contained in
1881365963 must have more than one figure in its root, because the
number is greater than 1000, which contains two figures in its root.
Then the root of 1881365963 may be regarded as made up of tens and
units; and, as the cube of the tens cannot give a less denomination
than thousands, the tens cannot be found in the last three figures, 963,
which may, therefore, be separated from the other figures. After the
separation of 963 the foregoing reasoning may be repeated, and thus
dividing the number into periods of three figures until we come, at
length, to the place occupied by the cube of the tens of the highest
order. The period on the left thus found, may contain but one figure,
as in the preset example, or it may contain two figures, or even three
^ures. We will designate the tens of the highest order by a% those
of the second order by a", those of the third by a"', &c. In like man-
ner, we will designate the units of the highest order by 2»^, those of the
second order by 6", &c.
EXTRACTION 01* BOOTS. 193
80^^=800= l8t Ap. diTltor. 1 881 8e6'»68'a25 1 2*8 *4 5
(80+2)2= 64=(8a^+y)y 1000
8a^+8aV+6^=3e4=Trtte divlior. 881 =(3a'«+8a'fe'+y»)5' +
84=:(8a'+y)ft' 728=(8 a^+8a^y+y»)y
4=ft^ 163 366=(3a"« +8a"y'+&"»)fc"+
8a"»=43200=2d Ap. diTlrar. 1828C7=(3a"«+3a"6«»+y'»)y'
(8flO+8)8=_10»=(3a''+y'>J/' 20 498 866=(8a"'«+3a'"ft^'+t'"«)ft*'+
44288=Tni« diTlsor. 182189M=(3o'"«+3a'"6"'+ft«"«)i"'
1089=(8a"+^'0^' 'i 285 060 625= (8a""»+3a""fe'w+6^''«)&*"
9=y"« 2 286 069 026=(3a""«+3a""y«'+&""^y"
8rt"'»=4688700=3d Ap. dJvUor.
(8090+4)4= _m76=(3a'"+V")6"'
4668476=Trne dlrisor.
14T76=(8a*'+d«0y
18=y ^'»
(3a<* )«=456826800=4th Ap. dlriBor
(87026+8)5 = 185128
457011925=Tme divisor.
In tills example, we begin by dividing the number off into periods
of three figures each ; the period on the left contains but one figure.
We extract the greatest cube contained in this period^ and set the root,
1, on the right. We then have found a', or the tens of the highest
denomination. We then bring down the next period. Now, since but
two periods are under consideration, the a', or 1, found, will be tens
with respect to these two periods, and therefore, three times its square,
or 3a'' sss 800. This is, then, the first approximate divisor. Dividing
881 by it, we get 2, or the supposed units contained in the root of the
two left hand periods, we then add 64, or (3a' -f b')I/, to the approxi-
mate divisor, and, if 1/ be the true units of the root^ 364, or Sa'^ -f-
3ai' -h i**, will be the true divisor. Now, since 364 enters into 881,
the same number of times that 300 does, 1/ has been found correctly.
We, therefore, multiply 364 by 2, and we form the three parts, (Sa^
^3a6' + &")&', of which the remainder, 881, is composed, except
some tens and units which have been incorporated in it from the cube
of the tens of a lower denomination. We subtract 728 from 881, and
bring down the next period. Now, the next approximate divisor is
plainly three times the square of 120, or 43200. But the shortest way
of getting three times the square of 120 is to add 64, and the square
of U, or 2, to the true divisor, and multiply the sum of the three by
100. The reason of this is apparent. The tens of the next denomina-
tion is found by dividing 153365 by 43200 j the quotient 3 is set on
the right of the 2 in the root, peparated from it by a dot. We next
add 1089, or (3a" + b")h", to 43200, and we have the second true
divisor. Multiplying this divisor by 3, and subtracting the product
from 153365, and bringing down the next period, we have a new
17 N
194 FORMATION OF THE POWERS AND
number, the root of whose unit is to be found. The approximate di-
visor to find h"' must be three times the square of 1230 ; and this is
most readily found by adding 1089 and the square of V, or 3, to
43200, and multiplying the sum by 100 ; that is, annexing two cyphers
to the sum. The approximate divisor thus found, 4538700, enters 4
times in 20498365, and the true divisor, when formed, enters the same
number of times ; 4 is then, truly, the fourth figure of the root. Mul-
tiply 4538700, the true divisor, by the last figure of the root, subtract
the product from 20498365, and bring down the next period. The
next approximate divisor is three times the square of 12340 ; this can
readily be formed like the preceding. The true divisor is formed when
5, the final unit, has been determined. The true divisor, multiplied
by 5, gives a product equal to the last remainder. Hence, the given
number is a perfect cube, and 12345 is its exact root.
The reason of the above process becomes very plain upon a slight
examination. It was established at the outset, that the tens and units
contained in the first two periods could be sought, independently of the
other periods. Having found found 1 ten and 2 units in the first two
periods, it is plain that this number, 12, is the tens in 123, the root
found in the first three periods. So, 123 may be regarded as the tens
of the first four periods. The root found in these periods is 1234,
made up of 123 tens and 4 units. In like manner, 1234 may be re-
garded as the tens of the root of the whole five periods. In fact, the
root found, 12345, is made up of 1234 tens and 5 units.
We then had tens of different denominations, 1 ten, 12 tens, 123
tens, and 1234 tens. To apply the general formula, (a + 6)* a=s a» +
3a'6 + 3(72i' + h^j for the cube of a number made up of tens and units,
to determining the tens and units contained in any two consecutiTe
periods, it became necessary to distinguish the tens by dashes to indi-
cate the denomination to which they belonged.
We will add another example, to show more fully the application of
the preceding principles. Required the cube root of 997002999.
3a'« = 24800 = App. divisor. 997 002 999
(3a' -f }/)V » 2511 729
26811 = True divisor. 268 002
2511 241 299
999
81 26 703 999
2940300= 2d App. divisor. 26703 999
(8a" + y')h" = 26811
2967111 = 2d True divisor.
EXTRACTION OP ROOTS. 195
In this example^ the quotient of the division of 268002, by the ap-
proximate divisor; 24300, is 10; bat this is manifestly absurd. By
trial, 9 is found to be the second figure of the root, because the true
divisor, 26811, enters 9 times, neither more nor less, in 268002.
We have for the extraction of the cube root of any number above
1000, the foUovnng
RULE.
I. Begin on the right and divide off periods of three figures each.
There wUl remain on the left a period of one y ttcoy or three figures.
n. Extract the greatest cube contained in the left hand period, and
set the root on the right, after the manner of a quotient in division.
Subtract the cube of the root from the left hand period, and brivg
dawn the next period,
m. Take three times the square of the root found, regarded as tens,
and set it on the left as an approximate divisor; see how often this
enters into the first remainder. The quotient will be the second figure
of the root, or something greater. Add to the approximate divisor iJie
product of three times the first figure of the root, regarded as tens,plw
the second figure of the root by the second figure of the root. The sum
of this product and the approximate divisor vnll be the true divisor
if it enter into the remainder the same number of times as the approxi-
mate divisor. MuUipHy the true divisor by the second figure of the
root, subtract the product from the first remainder, and bring doum the
next period.
TV. Add to the first true divisor the same number that was before
added to the approximate divisor, plus the square of the second figure
of the root, and annex two cyphers to the sum, the remit wiU be the
second approximate divisor. The true divisor is found as before.
Continue this process until all the periods are brought down; then, if
the last true divisor, multiplied by the final units of the root f gives a
prodtict exactfy equal to the last remainder, the given number is a per-
fect cube, and its exact root has been found.
EXAMPLES.
1. Required the cube root of 9261. Ans. 21.
2. Required the cube root of 85184. Ans. 44.
3. Required the cube root of 8024024008. Ans. 2002.
4. Required the cube root of 1371330631. Ans. 1111.
196 FORMATION OF THE P0WEB8 AND
5. Required the cube root of 1030607060301. Am. 10101.
6. Required the cube root of 1367631. Ans. 111. •
7. Required the cube root of 1879080904. Ans. 1234.
8. Required the cube root of 95306219005752. Ans, 45678.
9. Required the cube root of 95306219005752000.
Ans. 456780.
10. Required the cube root of 468373331096. Am. 7766.
APPROXIMATE ROOTS OP INCOMMENSURABLE NUMBERS.
264. Let a represent any incommensurable number — ^we have seen that
its root cannot be expressed by an exact fraction ; it may, howeveri be
truly determined to within less than any fraction, — , whose numerator
ar^ 7^ (r -\- 1)'
is unity. For a a= — 5-, and a > -g, and <^ ^^ — g— ^ ; r denoting the
entire part of the root of an^. Then, since a is comprised between
-=■, and ^5^ ~^, its root will be greater than — , and less than .
r . 1
Hence, — differs from the true root by a quantity less than — . Now,
as — may be made indefinitely small, the approximate root may be
found as near to the true root as we please ; the difficulty of extracting
the root increasing, however, with the increase of «.
Hence, we deduce the following
RULE.
Multiply and divide the number by the cube o/the denominator of
ihefradiion that marks the decree 0/ approximation ; extract the root
of the new numerator to within the nearest unit, and divide the result
by the root of the new denominator, which wiU be the denominator of
ihe fraction that determines the degree of approximation .
EXAMPLES.
1. Required the cube root of 4 to within \. Ans, |.
8/4 5* 8/500
Because yT= 4/ -^ = \ / -^ < f, and > J. The true
Ans.
y
Ans,
^ nearly
Am.
V
A 718,
y or
'#
Ans.
y
Ans»
V
BXTBACTION OF BOOTS. 197
root lies, then, between | and f , bat is nearer to f than to |. Hence,
I is the true root to within less than ^.
2. Keqxdred the cnbe root of 80 to within j^^. Ans, f ^.
^ ,,^ V«0 . 12» 8/138240 ^ ^, ^ _„
Because yW= ^-^^^ = V "12^ > '^ "^ "^*
3. Required the cube root of 90 to within j.
4. Required the cube root of 712 to within j,
5. Required the cube root of 1820 to within ^.
6. Required the cube loot of 200 to within ^.
7. Required the cube root of 2397 to within \.
8. Required the cube root of 1531 to within ^.
9. Required the cube root of 3575 to within J. Ans. ^
10. Required the cube root of 8375 to within ^.
Ans, Y; or 15, commensurable
11. Required the cube root of 1062208 to within ^. Ans. ^*
CUBE ROOT OP FRACTIONS.
265. Since the cube, or third power, of a fraction is formed by
raising the numerator and denominator separately to the third power,
the cube root of a fraction can plainly be found by extracting the root
of the numerator and denominator separately. The root of the nume-
rator written over the root of the denominator will then be the root of
the fraction.
There are three cases : 1. The numerator and denominator of the
given fraction may be both perfect cubes, and then the root of the one
written over the root of the other will be the required root. 2. The
numerator may be an imperfect cube, and the denominator a perfect
cube, then the root of the numerator extracted to within the nearest unit,
written over the exact root of the denominator, will be the approximate
root to within less than unity, divided by the root of the denominator.
3. Both the numerator and denominator may be imperfect cubes, or
the denominator only may be an imperfect cube, then the dcDominator
must be made a perfect cube by multiplying it by its second power.
The numerator must also be multiplied by the same number, otherwise
17*
198 FORMATION OF THE POWERS AND
the value of the fraction will bo altered. The root of the new nume-
rator, to within the nearest unit written over the exact root of the new
denominator, will be the approximate root required. If a nearer degree
of approximation be required, both terms of the fraction may be multi-
plied by the 5th^ 8th, &c. power of the denominator. The reason for
making the denominator rational rather than the numerator, is the
same as that given in the explanation of the principles involved in ex-
tracting the square root of fractions.
The rule for the extraction of the cube root of fractions belonging to
either of the three foregoing classes, is as follows :
RULE.
Make the denominator rational, if not already so, hy multiplying
hoik terms of the fraction hy the square power of the denominator.
Extract the root of the new numerator to within the nearest unit, and
write the root found over the root of the new denominator, which wHl
be the same as the denominator of the given fraction. Jf a nearer de-
gree of approximation he required, both terms of the fraction may be
multiplied by the t>th, Sth, <S^c, powers of the denominator. The root
of the new fraction will be the root required.
EXAMPLES.
1. Required the cube root of ^, An$, |.
For (|)» = I X I X f = ,ft.
2. Bequired the cube root of |j} to within \, Ans, j.
Because (|)' =r ^^, and (|)' = ^■^^, Hence, the true root lies be-
tween j and j ; and \, therefore, differs from the true root by a quan-
tity less than \.
8. Bequired the cube root of f to within \. Ant. \.
3/2 5«
Because vy= \J g^ = ^ || > |, and < f The true root
is nearer | than |, and, therefore, \ is taken.
4. Bequired the cube root of f to within i. Ans. f .
Because V'^^ \/^ = ^^> i and < f
EXTBACTION OF BOOTS. 199
' 5. Eeqmred the cube root of f to within y^^. Am, |^§.
6. Required the cube root of y to within J. Ans. 'g*.
266. The cube root of a fraction may also be determined to within
unity, divided by any power of the denominator; we have only to mul-
tiply both terms of the given fraction by such a number as will make the
denominator a perfect third power of the denominator of the fraction
that marks the degree of approximation, and then extract the root of
the new fraction. Let it be required to extract the cube root of ^^ to
within I, then the given fraction must have its numerator and denomi-
nator multiplied by 82, in order to make the denominator 64, the cube
of 4, the denominator of the fraction that marks the degree of ap-
proximation.
Then, yY = V^^^^ = ^"W < h and > f
8. Required the cube root of ^f to within J. Ans. g.
For y y = y/%^ = nn > §. and < §.
APPROXIMATE ROOT TO WITHIN A CERTAIN DECIMAL.
267. There are two cases: the given number whose root is to be
found may be entire, or it may be mixed ; partly entire and partly
decimal.
CASE L
Approximate root of whole numbers to within a certain decimal.
If the decimal fraction that marks the degree of approximation be
clianged into an equivalent vulgar fraction, the cube of its denominator
will be unity, followed by three, six, nine, twelve, or some multiple of
three cyphers. In other words, the cube of the denominator will con-
tain unity, followed by as many periods of three cyphers each, counting
from the right, as there are decimal places in the fraction of approxi-
mation. Thus (1/ = {^^y = ^5^5 • (-Ol)' = (riu)' =- TTHsijuo-
Then, to multiply the given number by the cube of the denominator of
the decimal changed into a vulgar Action is nothing more than an-
nexing three, six, nine, or some multiple of three cyphers. After a
new number has thus been formed, the extraction of the root is, of
200 FORMATION OF THE POWEBB AND
course, performed just as when the fraction of approximation was a
vulgar fraction. Let it be required to extract the cube root of 5 to
within -01.
rri, 3,T- ®/^ ■ (100)» 8/5000000 . , _ ^ ^ ,Ti
Then, y 5 = ^/ -^^^ = \/ -Jj^^^ > Hi. and < jJJ.
RULE.
Annex to the given number three times as many cyphers as there are
decimal places required in the root Extract the root of the new num-
ber thus formed to tcithin the nearest unity and point off from the right
the required number of decimal places ; which amounts to the same
thing as dividing the root of the new numbei* by the denominator of the
fraction of approximation changed into an eouivalent vulgar fraction.
EXAMPLES.
1. Required the cube root of 60 to within •! An^, 3-9.
2. Required the cube root of 1775 to within •!. Ans, 121.
«
3. Required the cube root of 9 to within *01. Ane, 2*08.
4. Required the cube root of 9 to within '0001.
Ans. 2 0801 nearly.
5. Required the cube root of 1864967 to within -1.
Ans. 1231.
CASE n.
268. Approximate roots of mixed numbers to within a certain
decimal.
Let it be required to extract the cube root of 2-3 to within *1.
Th.., VW - ^ = ^-^^ - ^^~ > „, . > 1.3.
We see that when the decimal has been changed into a vulgar frac-
tion, and the denominator made rational, the numerator, 2300, is the
given number, with the point omitted, and with cyphers enough an-
nexed to make the number of places, counting from where the point
was, a multiple of three. If it had been required to extract the cube root
of 2-3 to within -01^ it would have been necessary to add five cyphers
EXTRACTION OF ROOTS. 201
to 3 ; the number of decimal places then would have been made a mul-
tiple of 8. NoWy in pointing off for decimals, it is plain that we point
off as many places for decimals as there are places required in ihe root.
For, if one decimal place be required in the root, the denominator of
the root of the equivalent vulgar fraction will be 10; if two places be
required, it will be 100, &c.
r
RULE.
Annex cyphers to the decimal part of the mixed number until there
are three places, if one place be required in the root; six places if two
be required in the root, dfc. Extract the root of the new number thus
formed cu a whole number, and point off from the right the required
number of decimals,
EXAMPLES.
1. Required the cube root of 212 to within -01. Ans. 1*28.
For, »/-M2 = vVH - .yWEM « . yllOOOO
or 1-28.
2. Required the cube root of 41 to within 1. Ans, 1*6.
3 8/41 X 10« 8/4100 ^ , « , ^ , -r
For, yiT« y7i = \/ ^Q3 ° V "105" > t*' *^^< H-
3. Required the cube root of 8-88 to within 01. Ans. 207.
4. Required the cube root of 68-64 to within -1. ^ Ans, 41.
6. Required the cube root of 1770-25 to within -1. Ans. 12*1.
6. Required the cube root of 1150-455 to within -1.
Ans, 10-4 nearly.
7. Required the cube root of 5011-125 to within -1. Ans, 17-1.
In the 4th and 5th examples, one cypher only had to be annexed.
In the last two examples, no annexation was required. But if the
root in the last two examples is to be determined within -01, then
three additional cyphers must be annexed ; if within -001, six addi-
tional cyphers, &c.
The reason for annexing cyphers until the decimal places can be
separated into periods of three figures, is evident; even without chang*
202 FORMATION 07 THE POWERB AND
ing the decimal into an equivalent vulgar fraction. For each period
of three figures must give one figure in the root; if, then, the decimal
places were not made multiples of three, when we come to point off
from the right the decimals would be mixed with the whole number.
Suppose we were required to find the root of 8-72 to within •!. Now,
8*72 is but little greater than 8, whose root is 2. Hence, the root of
8*72 ought to be but little greater than two; but if we annexed no
cypler to 72, we would have but one period, 872, and the root would
be 9 approximatively.
APPROXIMATE ROOT OF DECIMAL FRACTIONS TO WITHIN
A CERTAIN DECIMAL.
269. A decimal fraction changed into an equivalent vulgar fraction
will have a rational denominator when its number of decimal places are
multiples of three.
Thus, .021 = jUv, -007681 = TJ^smj. 123456789 = TVtfWVbm-
Hence, if cyphers be annexed to the decimal until the number of
places are made multiples of three, the new fraction, when changed
into an equivalent vulgar fraction, will have a rational denominator.
And, since every period of three figures gives one figure in the root,
cyphers must be annexed until the number of periods of three figures'
is exactly equal to the number of places required in the root.
RULE.
Annex cyphers to the given decimal until it can he divided off into
as many periods of three figures each as there are places required in
the root. Extract the root of the new decimal thus formed to within
the nearest unit, and point off from the right the required number of
decimal pkices.
EXAMPLES.
1. Required the cube root of "8 to within *1. Ans. "9.
For, V^^y-J^ = ^^- > ^, and < jg.
2. Required the cube root of -08 to within -1. Ans, -4.
For, VW= V^= VT!]f7> tV "d < V\,.
EXTRACTION OF BOOTS. 208
3. Required the cube root of '008 to within -1.
Ans. -2, commensurable.
For, 4^008 = V^= i\ = -2
4. Required the cube root of -8 to within -01. Ans. •92.
5. Required the cube root of '8 to within *001. Ans. *928.
6. Required the cube root of -68 to within '1. Ans. *8.
7. Required the cube root of -68 to within *01. Ans. '87.
8. Required the cube root of -068 to within -1. Ans. '4.
9. Required the cube root of -0999 to within -01. Ans. '46.
10. Required the cube root of -999 to within -01. Ans. -99.
11. Required the cube root of '0125 to within -01. Ans. '23.
12. Required the cube root of -125. Ans. '5, commensurable.
In some of the examples^ it was necessary to annex one cypher, in
others, two cyphers ; in others, again, three cyphers. It will be seen
that the root is greater than the number; this ought to be so, since the
denominator of the decimal changed into a vulgar fraction is greater
than the numerator.
APPROXIMATE ROOT OF VULGAR FRACTIONS TO WITHIN A
CERTAIN DECIMAL.
270. Any vulgar fraction may be changed into an equivalent deci-
mal fraction by annexing cjrphers to the numerator, and dividing the
new numerator thus formed by the denominator. The equivalent deci-
mal will be mixed, or purely decimal, according as the given fraction
ifl improper or proper. The division of the new numerator by the
given denominator must be continued until the decimal part of the
quotient contains as many periods of three figures as there are places
required in the root.
RULE.
•
Change the vulgar fraction into an equivalent decimal fraction, and
make the d^mal part of the quotient contain a< many periods of three
figures as there are places required in the root. Then extract the root
according to the preceding rules.
204 FORMATION OF THB POWKBB AND
y
EXAMPLES.
1. Kequired the cube root of ^ to within •!. Ans. -6.
2. Hequired the cube root of | to within -1. Ans, 1-1.
8. Required the cube root of | to within -Ol* Ans, 1-17 nearly.
4. Required the cube root of | to within *01. An4, -82.
5. Required the cube root of -^^ to within *001. Ans, *4d6.
General Remarks on the Extraction of the Cube Root.
271. In extracting the cube root, we formed the three parts of
which each sucQessive remainder was composed, regard being had to
the denomination of the tens and units in those remainders. But,
since any number, when cubed, gives as many periods of three figures
each, counting from the right, as there are places of figures in the
given number (the period on the left, however, not necessarily contain-
ing more than one or two figures), it is plain that the first two periods
on the left of the given number give the first two figures on the left of
the root. The first three periods on the left give, in like manner, the
first three figures on the left of the root, and so on. It is evident, then,
that it is not necessary to form the three parts of which the successive
remainders are composed; we havo only, when the first figure of the root
has been found, to cube it, and subtract the cube from the left hand
period. When the second figure of the root has been found by using
the approximate divisor, 3a'', we have only to cube the two figures
found, and subtract the cube from the two left hand periods, and bring
down the next period, and so on. If the cube of the first two figures
of the root exceed the first two periods on the left, the second figure of
the root must be diminished until the cube of the two figures is equal
to, or less than the two periods on the left. The successive figures of
the root can only then be ascertained by trial. We can always tell, by
cubing the two, three, or four figures of the root found, when the last
figure is too great ; but some test is necessary to point out when we
have diminished the last figure too much. That test depends upon the
principle, that the difference between two consecutive numbers is equal
to three times the square of the smaller number, plus three times the
smaller number, plus unity.
Let a be the smaller number, then the consecutive number next
EXTKAOTION OF ROOTS. 205
above it will bo a+ 1. And (a+l)'— (a)» =» a»+8a"+3a+l — a*
= 3a' + 3a + 1, 80 enanciated. If, then, after subtracting the cube
of the fignres found ^m the periods on the left, the remainder is ex-
actly equal to three times the square of the root found, plus three times
this root, plus unity, the last figure of the root can be increased exactly
by unity. If the remainder is greater than this quantity, the last
figure can be increased by unity, or something more than unity.
The following examples will illustrate the foregoing process. Re-
quired the cube root of 531441.
531441
8a' = 19200 512
a + 2i
8'1
19 441
531 441 « (a + by.
After finding the tens of the root, the approximate divisor entered
once in the remainder, the quotient was set on the right of the root
found, and the two figures of the root (81), when cubed, gave the
original number. Hence, the number was a perfect cube, and 81 its
exact root.
Take, as a second example, 681472.
681 472
3a» = 19200 512
a fb
8 8
169 472
681 472 = (a -f&)^
The approximate divisor gives 9 for the second figure of the root;
but 89, when cubed, exceeds the given number, 681472. The last
figure, must, therefore, be diminished, if we make this figure, 7, and
cube 87, the cube when subtracted from 681472, leaves a remainder
exactly equal to three times the square of 87, plus three times 87, plus
unity. The last figure, then, can be increased exactly by unity, and
88 Ls the true root.
272. The formula for the difference between consecutive cubes,
enables us to pass from the cube of one nxunber to the cube of the
number next above it without actually cubing the higher number.
Thus, since, the cube of 10 is 1000, the cube of 11 must be 1000 -f
3 (10)» + 3 (10) +1 = 1000 + 300 + 30 + 1 = 1331. In like man-
ner, since (64)« = 262144 ; then (65)> = 262144 -f 3 (64)« + 3 (64)
+ 1 = 262144 + 12288 -f- 192 + 1 «= 274625. AVhen the numbers
18
206 FORMATION OF THK POWERS AND
under consideration are very large, the formula will save a great deal
of trouble in deducing the cube of the bigber number.
The second process that we have given for extracting the cube root
is usually followed, and when the given number contains but two
periods, is shorter than the first process ; but in every other case is
longer.
EXTRACTION OF THE SQUARE ROOT OF ALGEBRAIC
QUANTITIES.
278. We will first show how to extract the root of monomials.
Since all rules for the extraction of roots must be founded upon the
rules for the formation of powers, we must first examine, in the present
case, how the power of a monomial is formed. Let P represent the
numerical coefficient of any monomial, and a" the literal part of the
monomial. Then, the monomial will be Pa" ; now, to square Pa", is
to multiply it once by itself. And Pa" X Pa" = Pa'". Hence, to
square a monomial, we have only to square the coefficient, and to
double the exponent of each literal factor. The square of 2a'c is, by
the rule, 4a*c'. The square of 6a~^(^d~^ is, by the rule, 25a"~Vrf^.
The square root of 4a*c' is then, of course, 2a'c; but — 2a'c will, when
multiplied by itself, give 4aV. Hence, the root of Aa*<^ may be either
+ 2a'c, or — 2a*c. So, in like manner, the root of 25a~*cV~^ may
be either + 5a~'o'e?-^, or — Sa^'c'c^^
RULE.
L Extract the square of the coefficient, and divide the exponent of
eavh literal /actor hy 2.
II. Write, after the root of the coefficient, each literal factor, vnth
its new exponent.
III. Affect the whole result with the double sign, db.
EXAMPLES.
1. Extract the square root of 16a*c~*(^". Am. ± 4a'c"'rf^.
2. Extract the square root of IGa'^ft^^c*'. Ans. ± 4a"fi*"f ».
3. Extract the square root of 144a*fc'^c'«. Atis, =fc 12a^6V.
4. Extract the square root of 36a*»fc'"c»d«. Ans, =b Mh^c'^d}',
EXTRACTION OF ROOTS. 207
6. Extract the square root of a^'b^JC^^*. Am. ± a^^h^c"**,
6. Extract the square root of 256 —^ — •
7. Extract the square root of -^rn- . , „ - .
25
8. Extract the square root of
a^s^' Ans, =h 5x~'y~'2r'.
9. Extract the square root of JC ■. . . i? ,
^ 25 Ans. =h b~^xyz.
10. Extract the square root of P^^M^^N-". Ans. =h P^M^jN"*.
11. Extract the square root of t, ,, , . . , „^ 7_j i
12. Extract the square root of — .,.^ . . .^-.^ , ,,
^ 400 Ans. lb (20)~'a"fe^"c.
13. Extract the square root of a-'A-^'tr"'. Ans. ± a-*6-**c-^«.
It is plain that a monomial will not be a perfect power when its co-
efficient is not a perfect square^ and when every exponent is not some
multiple of 2. But when the exponents of the literal factors are not
exactly divisible by the index of the root, the division can be indicated.
1 i
Thus, v/4x = zfc 2x^, for, by the rules for multiplication, (zfc 2x^)
(± 2x^) = Ax. Hence, db 2x^ is, truly, the square root of 4a;. So,
also, yTaH^ = dz aH^c^, because (± ah^c^') (± aH^c^) = ah*c^.
The square roots, then, of all algebraic quantities may be truly ex-
pressed whenever their coefficients are perfect squares.
SQUARE ROOT OF POLYNOMIALS.
274. A trinomial is the least polynomial that is a perfect square.
It would be a mistake to suppose that the square root of a' + h^ is
a + 6, for (a + by = a' -f 2ab + ¥. The term, 2ab, enters into
the square of (a -f b\ and is not found in a' + 6*. Any polynomial,
to be a perfect square, must be susceptible of resolution into two equal
factors; and we know that when these factors are multiplied together to
reproduce the polynomial, the two extreme terms of the product (if the
208 FORMATION OF THE POWERS AND
factors have been arranged with reference to a certain letter) are irredu-
cible with the other terms. Hence, the square root of these extreme
terms must be terms of the whole root. The extreme terms of y/a*-^h'
must then be a and h, but the intermediate terms can only be found
approximatively.
Any binomial, as (a'+«), when squared, will give a trinomial, a'^-^
2a'« + «^. Conversely, if we have any trinomial that is a perfect
square, its root must be a binomial.
Let it be required to extract the square root of ISab + 81a' + h*.
If this trinomial be arranged with reference to the highest power of
one of its letters, as a, and the square root of the first term be taken,
we know that we have certainly one term of the root required. Be-
cause, from what has been said, we know that the first term of the
arranged trinomial is the product of the first terms of the two equal
factors into which the trinomial can be resolved. In other words, it is
the square of the first term of the required root. The arranged poly-
nomial is 81a' -I- 18a'*
81a' + ISab + h^ 9a + b + h\ We begin by
81a' = a"
18a -I- 6 = 2a'« + s extracting the root of
2a'8 -f s' = 18a& -f ^' the first term, and set
18a& + 1/ this root in the same
horizontal line with the
polynomial, and on its right separated by a vertical bar. We subtract
from the given quantity the square of the first term of the root ; there
remain, then, only the two terms, 18a6 -f ^"- Now, we know that
the first term of this remainder, ISah, is the double product of the first
term of the root by the unknown second term. If, then, we divide
18a6 by 2 (9a), or 18a, the quotient, ft, must be the second term of
the root. The root is then completely known. The result can be
verified by squaring the root 9a -f 6; or, since the remainder 18«Z»
4- i' corresponds to 2a'» -h «', and since 18a corresponds to 2a', if we
write 18a below the root, and b, which corresponds to s, on its rights
and then multiply 18a -h 6 by 6, we must evidently form the remain-
der, 2a'a -f- «*.
275. Since, when a trinomial is a perfect square, its extreme terms
must be perfect squares, and its mean term must be the double product
of the roots of these terms, we can tell in a moment when a trinomial
is a perfect square ; the mean term of the arranged trinomial must always
be the double product of the roots of the extreme terms. -Let us apply
BXTBAOTION OF BOOTS. 209
ihia rimple test to some ezpressioiis. 4a' + 4ma + m' is a perfect
aqiiarei and its loot 2a + m; a + 2^/ab + & is a perfect square, since
the test is satisfied, and the root, ^a -{• y/b; 7? +22;*y*+y' is a per-
fect sqnare, and its root, x^ + y^.
The root, however, will onlj be commensurable when the extreme
terms of the arranged trinomial arc rational. We have, from the fore-
going, a simple rule for the extraction of the root of a trinomial that is
a perfect square.
RULE.
Extract the root of the extreme terms, and connect their roots together
h^f the sign of the mean term.
Thus, the square root of wi* — 2mn + n" is wi — n ; this is obvious,
since (m — nf =s m' — 2m» -f n"; or it may be seen by going through
the steps of the process described.
EXAMPLES.
1. Required >/49a' + 14am + m'. Ant. la + w.
2. Required ^49a^ — 14am + m*. An$. 7a — m.
3. Required %/49a' + 28am + 4m". Ans. la -f 2m.
4. Required >/49a» — 28am + 4m». Ans. la — 2m.
5. Required v'4m + 16>/mn + ISn. Ans, 2>/m + 4V».
6. Required V 4m — 16\/m/i + 16ft. Am. 2y/m — 4Vn.
7. Required ^m^ — 14am + 49a'. Ant. m — la.
8. Required V 4m" — 28am + 49a'. Ans. 2m — la.
Remarks.
Examples 2 and 4, in connection with 7 and 8, show, that when the
mean term of the trinomial is negative there will be two distinct roots,
according to the arrangement of the terms. The reason of this is
plain. It is evident, moreover, that when either of the extreme terms
has the negative sign, or when both are negative, the root will be
imaginary.
18* o
210 FORMATION OV THS POWSBS AND
The role for the extraotion of the square root of a trinomial has an
important application in the solution of complete equations of the second
degree^ and ought, therefore, to be remembered.
276. If (he given number^ instead of being a trinomial^ liave a tri-
nomial root
Let a -{■ m + n represent this root. Then, by representing a + fti
by py and the given polynomial by N, we have N = (|> + n)* «
j9* + 2pn + n* =z o^ •\- 2am + m" + 2 (a + m) n + n*. We sec
that the first three terms is a perfect sqTiare, and the root may be
found by the rule for the extraction of the root of a trinomial ; then,
when we have taken the square of the root, a' + 2ani + m\ from the
given polynomial, there will remain 2 (a -f m) n + n*. The first term
of this remainder, 2an, divided by 2a, the double of the first term of
the root will give n, the third term of the root. The remainder,
2 (a + m')n + n", may be put under the form (2a -f 2m + n)n. If,
then, the first two terms of the root found be doubled, and the third
term added to them, and their sum be multiplied by the third term,
the product thus formed will be equal to the remainder. The process
for extracting the root of a polynomial which is the square of a trino-
mial, is precisely like that for extracting the root of a polynomial which
is the square of a binomial. The divisor, to get any tehn of the root
after the first, is twice the first term of the root ; the terms of the root
preceding the last term found are doubled, the last term added, and
the product of the whole, by the last term, subtracted from the succes-
sive remainders.
The different steps can be best exhibited by the following example :
4n« +12m?i +9m" +4n6-f.6m5+ft^
a« = 4n«
2am+ m'+2 (a -fm)»+n"=12mn + 9m»+4n6+6iw6+y
2aw+m"=12mn+9m»
2n+3m+5
4n + 3nt
4n+6m+6
2 (a+m)n4-n*==4n6+6mf+5*'
(2a+ 2fn+n)n=4n6+6m5+i»
We began by extracting the root of the first term, the root found
was set on the right, and its square subtracted from the given polyno-
mial. We had, then, taken out a' of the formula from the given ex-
pressions ; we next doubled the root found, and used this double root
(2a of the formula) as a divisor to find the second term. This, when
found (m of the formula), was set on the right of 2n, and connected by
XZT&AOTION OF KOOTS. 211
its appropriate sign. The double of the first term, and the second
term, written directly under the root, were next multiplied by the
second term. We thus formed 2am -|- m' of the formula, which, when
subtracted from the first remainder, left 4n5 -f- 6m5 + ^*> correspond-
ing to 2 (a -f- m)n + n' of the formula. The first term of this remain-
der (2an of the formula), divided by twice the first term (2a of the
formula), obviously, gave the third term (n) of the formula. The first
two terms of the root were next doubled, and the third term added to
their sum. Having thus formed 2a + 2m -{-n of the formula, we
multiplied this sum, 4n -f- 6m+&, by the third term, 6, (or n of the
formula). We thus plainly formed the three parts of which the re-
mainder was composed; and the product being exactly equal to the
last remainder, the polynomial had an exact root, 2n -f 3m -f- b.
The foregoing reasoning can be readily extended to a polynomial
whose root contains four terms. For, let N represent the polynomial, l,
the sum of the first three terms of the root, and n the last term of root.
Then, N = (? + n)* = ? -f 2/n + w'. Suppose the first three terms
to be a, & and c; then, N = (a + Z» -f- c)' + 2 (a -f 6 + c)» + n\
NoW; the first three terms is the square of a trinomial, and the root can,
therefore, be extracted precisely as in the foregoing example. After
the square of the root has been subtracted from the given polynomial
there will remain 2 (a + h + c)n+n*. The divisor to find n is, ob-
viously, still 2a, twice the first term of the root. And, since the
remainder can be placed under the form of \2(a+ b + c) + n \nj it
is plain that, if the last term be added to twice the sum of the first three
terms, and the whole sum be multiplied by the last term, we will form
the three parts of the remainder. The process for extracting the root
of a polynomial which is the square of four terms, is then identical with
that for extracting the root of a polynomial which is the square of three
terms, and that, we have seen, is the same as the process for extracting
the root of a polynomial that is the square of a binomial. Now, if the
root is composed of five terms, after the square of the first four terms
has been subtracted from the given polynomial, the remainder will be
found to be the double product of the four terms found by the unknown
fifth term^ and the divisor to find this term will still be twice the
first term of the root. And so the reasoning may be extended to a
polynomial whose root is made up of six, seven, or any number of
terms. Hence, for extracting the root of any polynomial, we have the
following
212 FORMATION OF THE POWERS AND
BUL£.
Arrange the polynomial with reference to one ofiu Utters. Extract
the root of the first term, and set the root in the same horizontal line
with the given polynomial, and on its right, separated from it hy a
vertical bar. Subtract the square of the root found from the given
polynomial, and bring down the remaining terms for the first remain-
der. Write double the first term of the root found immediately be-
neath the place of the root, divide the first term of the first remainder by
it, and write the quotient, which is the second term of the root, on the right
of the first term of the root, and also beneath and on the right of double
this tei-m. Multiply the double of the first term, and the second term,
itself affected with its proper sign, by the second term, and subtract
the binomial product from this first remainder, and bring down the re-
mxziriing terms for the second remainder. Divide the first term of the
second remainder by the double of the first term of the root, and the
quotient, affected with its proper sign, will be the third term of the root.
Set this third term in the root, and also in another horizontal line on
the right of double the sum of the first ttco term^ of the root. Multiply
the three terms in this horizontal line by the third term, and subtract
their product from the second remainder. Continue this process until
the final remainder is zero, the root will then be exact; or continue
until the letter, according to which the polynomial has been arranged,
has disappeared from one of the remainders. Then, if all the expo-
nents of that letter in the given polynomial are positive, we conclude
that the polynomial is not a perfect square.
Required the square root of 4x' + 12ary + 4xz + IQxl + 9^ + 6y«
+ 24y? + 8fe + 16Z«.
The polynomial is already arranged.
4z«+ 12ay+4«+16x/+9y«+6yH-24yH-8M-16^«'
4x«
1st Rem. =122y4-4a;24.162/+V+%2+24y^f 8&+16P-f«>
12xy 4-V
2z
4x-f3y
4r+6sr-f-«
4z+6y+2j+4l
2d Rem. z^4xz-\-\Qxl 4.62^24.24^^.8(0+ 16^«s
4a» +%* •'
3d Rem. »16z/ +24y^h8/«+16Z*
16xZ +24W+8^^16P
BXTBAOTION OV BOOTB. 213
277. After ihe beginner has become familiar with tbe preceding
principles^ it will not always be necessary to go througb tbe process
of forming the snccessive remainders. The successive products, distin-
guished by parentheses, may be all formed, and their sum taken at
once finom the given polynomial, as in the following example.
m* — 6mii H- 9n» + 4mp — 2mq — 12np + 6n^ + Ap^—ipq + g*
(m*) 4- »(— fann + 9n») -f *(4wjp — 12np + 4pa) + *(— *imq + gn<? —
n n n n
H>n^<f)
m — 3n4-2p--^
2m^-3n
2)11— dn+2p
2m— e»+42>— 9
We have distinguished the successive products by parentheses,
affected by exponents written on the left.
278. It is obvious that the successive steps required by the general
rule for extracting the square root of any polynomial, amount to nothing
more than subtracting the square of the algebraic sum of the terms of
the root, as they are found, from the given polynomial. In some cases,
then, it may save trouble to subtract the square of the algebraic sum
of the first two terms of the root from the given polynomial, then bring
down the remaining terms and find the third term. Next, subtract
the square of the algebraic sum of the first three terms of the root from
the given polynomial; and continue in this way until all the terms of
the root are found.
EXAMPLES.
1. Required \/x' + 4^ — 6xz + 4xy — \2zy + 9z'.
AnB. X +2y — Zz,
2. Required V x* + y* — 2aV + 4^* + ^t^j? — 42y .
Ans, a? — ^ + 2^.
8. Required k/ m" -f- ^n* — m» + -q — \xn + |xm.
Ans, m — 3» + ja;.
4. Required Va* — 2ax + x* + 2aw — 2an — 2xm -f 2xn + m*
— 2mn + n" Arts, a — x + m — n.
5. Required >/»• + y* + 2ie*m -f 2xy — 2a?n + 2yw — 2t^n +
f»« — 2mn + «*. An9. a^ '\- r^ -{-m — n.
214 FOBMATIOK OF THS POWEBS AND
6. Required w a"+ar+2ax+a»i+a«+xn+xw4- x+x**
2
. m n
7. Required \/sia^ + 9xy-{'^+9xZ'\-9xm + y + ^ + J +
2 -h 4 . ^n*. y^ + "2 + 2 ^ 2 •
8. Required </ ~ + ^ + 5|^ + 6my + 6mx + bzy + 5zx +
-4n«. — 4. ^ + 6m + 62:.
9. Required v'y" + 4xy -f 4a:* + 1 + 4x + 2y.
Ans, y + 2j5 + 1.
10. Required Va;* + lOxy + 4a^ + 12xy* + 9y*.
11. Required ^a^ + 2x^y + 2x*y + 2ar*y^ + ay + ccy.
-4n«. a;* + ary + a:^.
12. Required the square root of a;^ + lOay + 4x'y4-12ay+9y*+
2x' + 6a:V + 1^^^ + Cy* + «■ + 2a:y + y*.
Am, 7? + 2xy + 3y* + a + y.
i^emarA;.
The short process^ indicated in Art. 277, cannot be followed in the
last example.
SQUARE ROOT OF A POLYNOMIAL INVOLVING NEGATIVE
EXPONENTS.
279. The principles for the extraction of the root of a polynomial
containing negative exponents are the same as for the extraction of the
square root of a polynomial, all of whose exponents are positive, observ-
ing, however, in the arrangement of the polynomial, that that negative
exponent is the least algebraically which is the greatest nmnericallj.
When, too, the arranged letter has disappeared from any remainder, it
SXTEAOTION OF BOOTS. 215
may be arapplied with a zero exponent, provided, that the exponent of
the same letter is negative in some of the terms.
Take, as an example, ar^ + 2x"'y + a^y + 23?^* +2y + y*.
Arranging the polynomial with reference to the ascending powers of
sc, and proceeding as before, we have
ar* + 2ar»y + 2x'^ + xY + 2xf + off
x-^
Ist Kem. =2x-V + 2a;V + ^y + 2xy« + xy
2g-> + xy
2d Rem. = 2x°y + 2xf + aV
2x°y + 2xy" + a;y
ar* + y +a;y
2x-» + y
2a;-" + 2y + ay
The firut term of the second remainder is 2y ; x, the letter accord-
ing to which the polynomial was arranged, does not enter into this re-
mainder until supplied. It, of course, must be introduced with a zero
exponent, otherwise the expression 2y would be altered by its introduc-
tion. But, since a;° = 1, then, 2xPy = 2y.
Take, as a second example, x"* + 2x~*jf + t/* + 2 -f- 2ya;" + aJ*.
x-^ -\- 2x-^y + y« + 2 + 2yx» -|- x^
^
1st Rem. = 2x-'*y -f y* + 2 + 2yx* + x*
2a;-V -f y'
2d Rem. = 2x» + 2t/x« + x*
2x° -h 2ya;* + x*
ar* + y + c^
2ar*-|-y
2ar« -f 2y + a:"
The first term of the second remainder is 2 ; a;, affected with a zero
exponent, was introduced into that term.
3. Required v/x"* + 2x-^tr» -h y^ + 2ar^ + 2x^ + x*.
Ans. X-* + y-* + a:.
4. Required the square root of -4-+ ^ 3 "^T"^^ "^""T^^*^"^'
^iw. — + Y + ^-
ar* x-V^ y~* 1 2xV^
5. Required the square root of -j- H ^ 1" q "^ "^ 3 — "*"
a^ + 2ar« -f ^ -|- 4x« -h 4. ^n<. ^ -j. ^ + 2 -I- x*.
216 FORMATION OF THE P0WEB6 AKD
6. Required the square root of ac* — 2a:*y + ajy + ^o*^' — 2x» +
2 — 2x-'^y + 2x-V' + ar« + y"*. -4«»- ^ — a^if + or* +y"'.
7. Required the square root of a;"^+2y4-xy H — j — f-aj'fy^'+l-
-4.n«. x~* H ^ 1- ary.
INCOMMENSURABLE POLYNOMIALS.
280. Wlien the exponent of the letter, according to which the poly-
nomial is arranged, is positive in all the terms, we know that the poly-
nomial is incommensurahle when this letter is not found in any of the
successive remainders, or is found affected with a lower exponent than
it appears with in the first term of the root. But, if the letter, accord-
ing to which the arrangement is made, appear in some of the terms
with a negative exponent, we can only tell that the given polynomial
is not commensurahle by observing that the operation would never end.
EXAMPLES.
1. Bequired the square root of a* + &'.
2. Required the square root of 1 — a:'.
3. Required the square root of re* — 1.
Ans, X : s— , — t-s-t + &C.
4. Required the square root of 2x' + 2a; -h 2.
13 S
Am, xy/T+ -^ + - — = + &6., or ^T(x + ^ + ^ + &c.)
V2 4a;v'2 o^
5. Required the square root of 2x^ — 6x + 4.
3 1 1
Aru. x^Y = ■=_ — &c., or VTCx — I — t. ftc.)
V2 4x^2 ^ ' 8x ^
6. Required the square root of m + n.
Aru, y/m+ —j=. — - =+ =.. — &C.
7. Required the square root of ar* -h 2a;-* + 4x-*
Ant. x-« + 1 + 2a; + &c
EXTEACTION 07 ROOTS. 217
SQUARE ROOT OP POLYNOMIALS CONTAINING TERMS
AFFECTED WITH FRACTIONAL EXPONENTS.
281. Sinoe^ in mnltipHeationy the exponents of like letters are added,
whether ihey be fiaotionali or entire^ it is evident that the square of
m m m 2m ta m
oF = a^ X a"" is oT, Hence, the square root of a^ is aF. Quanti-
ties involving fractional exponents may then be operated upon in the
same manner as quantities containing only entire exponents. This will
be shown more fully under the head of fractional exponents. Assum-
ing a truth which scarcely needs a demonstration, we will give a few
examples of polynomials involving fractional exponents.
EXAMPLES.
1. Required j^/x"^ -♦- y* — 2xy — 2y»z* + 2x'z^ + z.
Am. a^ — y*-|- 2 .
2. Kequired s/ ^ + ^ + ^^ + 1. + ^^ + ^J^.
i i i
. X y z
^~- T + T+ 2-
3. Required V-9 + 9 + -9-+9+-9-+ -9-
i i i
^"'- y + T+F
4. Required y/^^ + y* + 2a; V^ + 2yTSa« 4. 2x^a^ + a\
11
AfiB. x^ + y *^ 4- «*•
CUBE ROOT OF POLYNOMIALS.
282. The cube of a monomial, as a, is a', for (of = a x a X a = o*.
The cube of a binomial, a + 6, is o* + 3a'6 + Sab^ + h\
For, by actual multiplication, we have (a + by = (a + i) (a + b)
(a + 2>) = a' 4- 3a'6 -|- SaJ' + &•. Since, then, the cube of a mo-
nomial is a monomial, and the cube of a binomial a polynomial of four
19
218 FOBMATIOM OF THS POWEBB AND
terms, it follows that a polynomial of four terms is the least polynomial
whicH can have an exact cube root.
Knowing tHe third power of a binomial, we have only to reverse the
process to obtain the cube root of the power. We see that the fint
term of the power is the perfect cube of a, and we know, from ihe man-
ner in which a product is formed, that that term, a', has been derived
without reduction from the multiplication of the three factors of which
the power is composed. If then we extract the cube root of a', we
know certainly that the root, a, is a term of the required root. Then a
will be one of the extreme terms of the root; and, since the extreme
terms of the root when cubed ^ve powers that are irreducible with the
other terms, it follows that the cube of a or a' is to be taken from the
given polynomial. After this subtraction, the remainder is 3a*& +
da5' + ^j <^nd it is plain that the second term of the root can be found
by dividing the first term of the arranged remainder by 3a', or by three
times the square of the first term of the root. The remainder, 8a% +
Sa^" + b* can be put under the form of (3a' + 3a& +h*)h. If then
we add three times the square of the first term of the root, three times
the first power of this term by the second term of the root, and the
square power of the second term of the root, together, and multiply the
sum of the three terms by the last term of the root, we will obviously
form the three parts of which the remainder is composed. The pro-
cess, then, for extracting the cube root, is analogous to that for extract-
ing the square root of a polynomial.
a» + 3a«6 + 3aJ" + h*
a»
1st Rem. = 8a«6 -f Safc* + b*
3a«6 + daV + P
2d Rem. =0
a + b
3a" + 3a6 + 6«
The root is set in the same horizontal line with the given polyno-
mial, the cube of the first term is taken from the given polynomial, and
the first term of the arranged remainder is divided by three times the
square of the first term of the root, to obtain the second term of the
root. Finally, immediately beneath the root is written, three times
the square of the first term of the root, plus three times the product
of the first and second terms of the root, plus the square of the second
term of the root, and the product of the sum of these three tenns bj
the last term of the root is taken from the first remainder.
EXTRACTION OF R00T8. 219
If the root be composed of three tenns^ o, h and c, we may repre-
sent the algebraic sum of a and b by m. Then, let P be the poly-
nomial whose root is a 4- 5 + c, we will have P s (a -f- 6 + c)* s
(w + c/ = m» + Sw'c -h 3mc? + c? — (a + &)' + 8(a + 6)«c + 3
(a + i») c« + c* = a» + 8a»Z» + 3ai* 4- 2»' + 3a*c -(• 6a6c + 36*c + c*.
The first four terms of this polynomial are the same as in the last ez-
ample, and therefore their root, a + ^9 o^^ be found as in that example.
After the cube of the sum of these terms has been taken from the given
polynomial^ the remainder may be placed under the form of I 3 (a + by
-f 3 (a 4* ^) <^+ ^ I <^* Then,. haying found the third term by dividing
(he firat term, So'c, of the arranged remainder, by 3a', it is plain that the
remainder itself can be formed as before, by multiplying the last term
found into three times the square of the sum of the other terms of the
root, plus three times the first power of the sum of the other terms into
the last term, plus the square of the last term.
Now, whatever may be the number of terms in the root, we may re-
present the algebraic sum of all of them, except the last, by «. Sup-
pose n to be the hist term. ThenP ss (« -f »)> = <* -f- 3«'n + 8sn'-h n':
in which t represents the algebraic sum of any number of terms. If
a is the first term of <, then the first term of the development of 3/»,
will be 3aSi. Hence, the divisor to find n Is still three times the
square of the first term. After «* has been subtracted from the poly-
nomial, the remainder can be put under the form of (St^ + 3m -h »")».
Hence, whatever the number of terms in the root, the successive re-
mainders can be formed just as when there are but two or three terms
in the root
The following examples will illustrate the process.
sfi
1st K.=:3«»+ax*+»aB*y+*"+18x"y+27xV+9*^+27«»*-i 27y»
2d Rem. = 9x^ + 18x*y + 27xV + ^^ + 27a!y»+ 27y*
SdRau. =00 00 00
(3x* + 8«c« 4- sfl)x
x^+ ^a*+ 15x*+ 20x'+ 15x«+ 6x+ 1
x«
Ist R.=6a*-h 15x*-f- 20x^4- 15x*-f 6x +1
o(^+2x-\-l
(3x*+6x»-|-4a:^)2x
(S{a^+ 2a;)«+ 3(x*+ 2a;)l+ 1)1=
(3a:«+ 12x»+12x"-F3x«+ Cx-f 1)1
2d Rem. = 8x*+ 12^*+ 15x»+ 6x + 1
8d Rem. = *0 OHB
220 FORMATION OF THS POWE&S AND
RULE.
Arrange the given poiynomial with reference to the atcending or de-
Kending powers of one of its letters. Extract the cube rooi of the term
on the leftf and set the root on the same horizontal line with the given
polf/nomial on the rights and separated by a vertical line. Suhtrad
the cube of this first term of the root from the given polynomial, and
bring down the remaining terms for the first remainder. Write three
times the square of the first term of the root immediaidy beneath the
place of the root. This will be the divisor to find aU the other
terms of the root. Divide the first term of the remainder by the divisor j
and set the quotient, with its appropriate sign, on the right of the first
term of the root, for the second term of the root. Set, on the right of
the divisor, three times the product of the first and second terms of the
root, affected with its proper sign, plus the square of the second term of
the root. Next, multiply the algebraic sum of the three terms, thus
formed, by the last term of the root, and subtract the product from the
first remainder. Bring down the remaining terms for the second re-
mainder, and divide the first term by the divisor. The quotient is the
third term of the root. Take three times the square of the sum of the
first two terms, and add to this three times the product of the algebraic
sum of the first two terms by the third term, plus the square of the third
term. Multiply the algebraic sum of the whole by the third term, and
subtract tJie product from the second remainder. Bring down the re-
maining terms from the third remainder. Find the fourth term of
the root as before, and continue the process until aU the terras of the root
are found.
Remarks,
1st. Wlien the exponent of the letter^ according to which the ar-
rangement is made, is positive in all the terms of the given poljnomiali
we know that the root is not exact, whenever the exponent of the as-
sumed letter is less in the first term of any of the successive arranged
remainders than it is in the divisor.
2d. It is evident that the steps of the process, according to the role,
amount to nothing more than suhtracting the cuhe of the algehndc sum
of the terms of the root, when found, from the given polynomial. It
may, therefore, sometimes save trouble to proceed thus. Subtract the cube
of the first term from the given polynomiaJ. Find tho second term of
BXTBAOTION OV BOOTS. 221
tlie root according to the rale. Cube the algebraic sum of the two
terms of the root, and aabtract the result from the given polynomial.
Find the third term, and subtract the cube of the algebraic sum of the
first three terms of the root from the given polynomial. Proceed in
this manner until we get a zero remainder, or until it becomes evident
that the given polynomial is incommensurable.
EXAMPLES.
1. Required the cube root of a:* -f- 3x» + 8aj' + 7a» + 12a» + ^*
+ 12a? + \22?+ 8. Ans. £c'+aJ»+2.
2. Required the cube root of a« + 6x» + 21a* + 44a;" + 63x«
+ 54a: + 27. Ant. a:" 4- 2x + 8.
3. Required the cube root of 8a:* + 48a:* + 168a:* + 362a:' + 504a:»
+ 432a: + 216. Am 27? + ^x + 6.
7? 7? 11a:* 28x* 11a:*
4. Required the cube root of -g- + x + -^ +"27"'*" ~6~"^^ "*" ^"
a: a:
^"*-2'*"3+^
5. Required the cube root of ^. - + —^^ + -^-r- + — .
o4 o4 d4 o4
- 5a:* la;
Am. — r- + -7".
4 4
AH • ^ *i. u. * 4. 125a:* . 75a:' ,15a;» . a:»
6. Required the cube root of —5 J- -q— + -5 — V -5-.
o 000
. 5a:* la;
^«'- T" + r
7? 3a;' 3.x' 7a:* 3x* 3a:*
7. Required the cube root of -5-+-^+-^ + -^- + -5- + -j-
0000 ^4
3ar» 3x" - , 7? a? ^
+ -2" + -g- + 1- -^^*- 2" "^ ■2 ■*■
8. Required the cube root of 8a* + 12a'6 — 12a»f» + 6a5' —
12amh + 6am' — 3m6' 4- 3to'6 -f 6* _ m' -4n«. 2a + & m.
9. Required the cube root of ar'— 9x'y — 3a;' + 27x5^ + 18x3^
+ 3x — 275^— 272/*— 9y — 1. ^»«- a; — 3y — 1.
10. Required the cube root of x' + 3a:' + 3xy — 3xV + Sx* + x*
+ 6xy— 6x»y + 6x»y + 3x'y — Gx'y*— 3a:V + ^^f — ^^ +
8xy* + 2/* — S^ + 3y*— y*. Am. 7? + x + t/*— y.
19*
222 FORMATION OF THB POWBBS AND
CUBE ROOT OF INCOMMENSURABLE POLYNOMIAL&
288. — EXABfPLSS.
1. Required the cube root of as* + Saj" + 8x + 3.
2
Arts, a: + 1 + ^-j-+,&c.
2. Required the cube root of a^ + 6a^ ^ 12x + 9.
Ans. X + 2 + ^-5-+, &c.
8. Required the cube root of x" + a'.
o» a»
AnB.x+^-^&c,
4. Required the cube root of a^ + 3aj* + 4.
X
6. Required the cube root of cc + Sx" + 6.
a^
Ans. yx + -7=+,&c.
6. Required the cube root of x"* 4- Sx"" + 3x~* 4- 2.
Ans. x~' + 1 + g — 1-+, Ac.
7. Required the cube root of x~* + 6x^ + 12ar* + 9.
Ans. X * + 2 +Q-z5-+> &c-
8. Required the cube root of x~* + 8x^ + 3x-* + 2.
Ans. x-^+1 +g_j.+,&c.
9. Required the cube root of x* + 9x* + 27x* + 26.
3x»
10. Required the cube root of x* 4. 8x* + 3x* + x*.
Ans. X* + 3 — 5-5 — , &c.
Ans. X* + X — o"~"> &^
XXXBAOTION or BOOTS. 228
CUBE KOOT OP POLYNOMIALS INVOLVING FRACTIONAL
EXPONENTS.
284. — EXAMPLES.
3 1
1. Bequired the cube root of x^ +Sx + 3ic' + 1.
Ans. x^ + 1.
2. Required the cube root of x + 3a;^ + 3a;» + x^
3. Required the cube root of x* + 3x* + x^ + 8xf
Ans, X* + x'.
61 12 8
4. Required the cube root of x** + 3x ' + 8x » + x*.
1
Ans. x* + x^.
6. Bequired the cube root of x' + 3x^ + 3x* + 3x' + 6x^ + 3x"
4.x^ + 3x + 3x^ + 1. Ans, x* + x^ + 1.
6. Required the cube root of 8x" + 24x^ + 24x* + 48x^ + 24x"
+ 8x^ + 24x + 24xi + 8. Ans. 2x« + 2x^ + 2.
7. Required the cube root of x* + 3x» + 3x • + x*.
Ans. X* -f- X.
8. Required the cube root of x^ + 3x2 + 3x« + x^^.
1 J7
Ans. x^ + x*».
9. Required the cube root of x* + 3x ^ + 3x ' + x».
Ans. x' + X*.
10. Required the cube root of x" + 3x^ + 8x^ + 1 + 8x^ + 3x^
+ 6x^ + 3x* + 3xi +xi Ans. al^ + x^ + 1.
11. Required the cube root of x"" + 3x"^' + 3x'^ + 1 + 3x » +
ex"^^ + 8x~i + Sar^ + x"^ + 3x"i. Ans. x-* + x"^ + 1.
224 FORMATION OV THI POWERS AND
12. Required the cube root of ar^ + 3x * + 3tc » + x ^
Ant. ar* + 5c"*-
13. Bequired the cube root of a:* — 3x » + 3x* — x.
Ans, of — % .
14. Required the cube root of »• + 3x* + 6x* + 3x^ + Gx^ +9xi
+ lOic* + 9x« + 7a;i + 6a: + 3ic^ + 1. Ant, a;* + a; + a^ + 1-
15. Required tbe cube root of 1 + 3aj-» + 6x-« + 3x~5 + 6 "^ +
9x""^ + lOx-* + 9ar^ +Jx'^ + 6ar* + 8x" «* + ar*.
-4n«. 1 + x-» + x"^ + af-*.
16. Required the cube root of 27x«+81x»+162x*+ 81x^+162x3 +
243x^ + 270x»+243x«+189x^ + 162x + 81x3 + 27.
Anz. 3 (x* + X + x^ + 1).
CUBE ROOT OF POLYNOMIALS CONTAINING ONE OR MORE
TERMS AFFECTED WITH NEGATIVE EXPONENTS.
285. The process is the same as for polynomials, all of whose terms
are affected with positive exponents; observing, in the arrangement,
that that negative exponent is the least algebraically which is the great-
est numerically, and that when the letter, according to which the
arrangement has been made, has disappeared from any of the succes-
sive remainders, it may be introduced affected with a zero exponent.
EXAMPLES.
1. Required the cube root of x~" -f 3 + 3x' + x*.
Ann, x~* 4- X*.
For; by the rule
a;-« -I- 3 + 3x« + x^
X"*
1st Rem. = 3x^ + 3x« + x"
3x^ + 3x' + x"
2d Rem. = 00
x-« + x*
(3x-^ + 3x" + X V
BXTBAOTION OF BOOTS. 225
This polynomial might have been arranged with reference to the de-
scending powers of x.
2. Required the cube root of a?"* + 8ar^ + 3 + x*.
'Ans. or* + X.
3. Required the cube root of x"* + Sx^ + 3x-* + 3ar* + 6ar* +
4 + 8x + Sac^ + a*. Ans. or* + x + 1.
4. Required the cube root of x* + 3x* -f 3x + T + 12x-' + 6x-* +
12ar^ + 12ar' + 8ar*. ^iw. x + 1 + 2ar«.
5. Required the cube root of 1 + 6a:-* + 21ar* + 44x-» + eSar* +
64ar^ + 27ar^. Ans. 1 + 2x-» + 8ar«.
6. Required the cube root of oH* — Qcr^b + 12a-«&» — 8a-^5».
J.n«. a""* — 2ar^b.
7. Required the cube root of x* + 6x» — 40 + 96x-* — 64x-«.
Ans, X 4- 2 — dx*"*.
8. Required the cube root of 1 + 6x-' — 40x-^ + QGx-* — 64x-^.
Ans, 1 + 2x"* — 4x-*.
9. Required the cube root of 1 — 6x^' + 15xr* — 20x-' + 15x-* +
— 6x-* 4- xr«. Ans. 1 — 2ar^ -f x-^.
QUANTITIES AFFECTED WITH FRACTIONAL EXPONENTS.
286. Suppose it be required to multiply a into itself until it is taken
1 X
three times as a factor. Then^ by the rules for multiplication^ a X a^
X a = a'. This result is, evidently, the cube of a^. Now, a*, mul-
tiplied by itself once, gives a^xa^ == d^ ssia; and >/a, multiplied by
itself once, gives y/aXy/a=^y/a^=a, So, a , and >/a, are equi-
valent expressions. The expression, a^, then, which indicates the cube
of ct y also indicates the cube of >/a. The numerator of the fractional
exponent denotes the power to which the quantity has been raised, and
tbe denominator the degree of the root to be extracted. Taking or
three times as a factor, we have arXcrXar ^ ^iHH _, ^ j^^^^
226 FOBMATION OF THE POWERS AND
the ^a, raised to the third power, will also plainly be a, because,
from the definition of Involution and Evolution, it is evident that
the latter undoes what the former has done. The root of any quantity
raised to a power indicated by the index of the root, must then be the
quantity itself. Hence, (^^ a)^ =: a ; then, a ss ya; since we have
seen that both a , and %/ a, when cubed, give a. In general, a^ =s
m
v^ a"". Because oa , taken as a factor n times, ^ves cm X a* X aa s
w I ni I IB
an u n ss a», and Va», raised to the n^^ power, is also a". The
fractional index may be regarded as denoting a potoer of a root The
denominator expresses the root, and the numerator the power. The
denominator shows into how many equal factors, or roots, the given
quantity is divided, and the numerator shows how many of these fac-
tors have been taken. The fractional index may be considered to de-
note the root of apouser, as well as the power of a root Thus, a* may
either denote that the third power of the *Ja has been formed, or that
the square root of the third power has to be taken. Taking the former
view, it indicates an executed operation ; taking the latter view, an
unexecuted operation.
m
In general, a» may be read the m^^ power of the n^^ root of a, or the
n*** root of the m^^ power of a. It is more convenient to read the ex-
pression thus, a io the m divided by n, power; or a, raised to a power
denoted by the quotient of m, divided by n. So, a ^ is read, a to the
P
~ power.
?
There are two consequences of notation by means of fractional expo-
nents that deserve consideration. 1st. Any multiple of the numerator
of a fractional exponent may be taken, provided an equal multiple oS.
the denominator is also taken. Thus, a=a^ is also =iar s:^ar=3a^j fte.
2 rp
So, also, a4 = cek. The reason of this is plain; the increase of the
denominator makes the equal roots or factors of the given quantity, a,
smaller; but the numerator being increased just as much as is the
denominator, the number of these smaller equal roots will be increased
enough to make up for their diminution in magnitude. 2d. Since the
fractional exponent may be exchanged for any other of equal value, it
may be expressed in decimals. Thus, a = a*^ = a**. So, also, a^
EXTBAGTION OP ROOTS. 227
These decimal indices are called logarithrM. The manner of calcu-
lating them; and their use, will be shown hereafter.
MULTIPLICATION OP QUANTITIES AFFECTED WITH ENTIRE
OR FRACTIONAL EXPONENTS. — MONOMIALS.
287. Quantitities with fractional exponents must plainly be multi-
plied in the same way as quantities affected with entire exponents.
That is, the exponents of the same letter or letters in the multiplicand
and multiplier must be added, and after the common letter or letters
must be written those not common, with their primitiye exponents.
Let it be required to multiply a* by a'. Then, a* is to be repeated
a*, or a' times. Now, to repeat a^, a^ times, we have only to add
the exponents of multiplicand and multiplier, for the first exponent
1 8
denotes that a^ has been taken three times to produce the product, a^;
and the second exponent denotes that a^ has been taken four times in
the product, a^. Hence, a^ must enter seven times in the result of
8 4 J7
the multiplication of a^ by a^, and that result must be written cfi. To
mtdliply cfl by &, is to repeat a^, h times. Hence, a^X & = d^h,
RULE.
3hdt{pfy the coefficients together for the coefficient of the product.
Reduce the ea^nents of the same letters in multiplicand and multiplier
to the same denominator j add their numerators, and set their sum over
the common denominator. Annex these letters to the new coefficientj
and vrrite afier them all the letters which are not common to the muUi-
pltcand and multiplier,
EXAMPLES.
-L J. 1, £ — -1
1. Multiply a»i» byo"»6»c. Ans, a^h^c.
2. Multiply — g — by — 3— • -^»*- — g — •
228
FORMATION OF THE POWERS AN
3. Multiply
. , ah^c^ , ah^<^
m
by
Ans,
4. Multiply ^ by aMy^zf
X y z
«! 1 1 i-
6. Multiply a~"6 "c p by a»6"cp .
6. Multiply a" W 2 by JbK^.
7. Multiply 2a"^6~'»"c" ■ by — ^^^ — .
Ant, a*
iin<.
8. Multiply Sx^y^z^ by — 7x*6*z*. Ans. — 66x'
9. Multiply 12a-6-c» by 5a*6*c*.
-Ana. 60a ■ b'
la r p « r p
10. Multiply 4a "^"^0 <i bya»6~c<J.
^1
DIVISION OP QUANTITIES AFFECTED WITH FRACTK
AND ENTIRE EXPONENTS. — MONOMIALS.
288. Since the exponents of tbe like letters are added in mi
tion, they must be subtracted in division. For, tbe object of
is to find a quantity called the quotient, which, when multipl
the divisor, will give the dividend. Let it be required to divide a* by
a ; then, we must find a quantity, which, when multiplied by a mil
give a*. This quantity is plainly a^, for a X a' = a ^ = a^ = a".
1
But the quantity found, a^, has resulted from the subtraction of the
exponent of a from 2 ss ^, the exponent of a'. So, the result of the
I I
1 1
division of a« by a»" must be a" ", since a»~»"x a"" = a". When
the denominators of the fractional exponents are difierent^ the subtrac-
tion can only be indicated, not performed, until they are reduced to
the same index. If the exponents of the same letter in the divisor
exceed that of the dividend,' the exponent of that letter in the quotient
£XTBA0TION OF ROOTS.
220
[e. So; also, if there are any letters in the divisor not
dividend, they mnst appear in the quotient, with their
ments taken with a contrary sign ; else, when the quo-
ior are multiplied together, these letters would enter in
RULE.
)ejffkient of the dividend hy that of the divisor y the result
7t€nt of the quotient. Write after it all letters common
id divisor, affected with exponents equal to the difference
lents in the dividend and divisor; and, also, all the
n to the dividend only, with their primitive exponents;
ion to the divisor only, with their primitive exponents
zontrary sign.
EXABCPLES.
4a«c V by 2a*c*d*.
1 1 i 1
12a'ft»cVby6a»6«.
I
ide Ix^y^z^ by x^y^z'^,
Kvide 7z y hy z y x .
Divide 24a^6*c* by 24z»a"^rV^.
i i i 18
6. Divide 18x y z hj
-i -i -t'
X y z *
7. Divide 3a"'&" ' c* by ia'^h ' ^T" .
Ans, 2a^cd
i.^"2V
Ans. 2a ■' 6 «• c cP.
Ans. 7z p .
Ans. 7z~^y~^x .
Ans. Z'^dbc.
Ans. 1.
Ans. db'c^d"^.
8. Divide SOm'^K^ by ^mh'^x^. Ans. l&m^Hc^x^^.
_i * L fp*-H) (i-H»> («-h»
9. Divide 360c -/"^^'^ hylSOefy. Ans. 2c - / ■ ^ p .
10. Divide 4a»c'6* by a'^d'^e .
20
Ans. ch^dKK
228
FORMATION OF THE POWERS AND
3. Multiply
. , Jbici , ah^<*
m
by
i
4. Multiply -TTT ^^ «*«M«^-
5. Multiply a"»5 "c V by a*b»ep .
-1 «L -1 i 4 i
6. Multiply a •& <ic ^ by a*6'c*.
7. Multiply 2a'^5~«"c"" ■ by — ^ — .
4 I i ._ . KJJ
9. Multiply 12a-6-c» by 5o*fe*c*.
ilfM.
t2l
8. Multiply Sar'y^z* by — 7x h z . Ans. — 6Qx^y*z
Ans. at
Ans*\
An$» a ^ b
I
Ans. a«"i
M r p M r p
10. Multiply 4a ■& ""c i bya»6~cfl.
^flS.
DIVISION OF QUANTITIES AFFECTED WITH FRACTION.
AND ENTIRE EXPONENTS. — MONOMIALS.
288. Since tbe exponents of the like letters are added in mull
tion^ they must be subtracted in division. For, the objeet of divi
is to find a quantity called the quotient, which, when mtiltiplied^
the divisor, will give the dividend. Let it be required to divide a* by
a ; then, we must find a quantity, which, when multiplied by a will
give o*. This quantity is plainly a^, for a x a^^a ^ = a^ ^= a".
1
But the quantity found, a^, has resulted from the subtraction of the
exponent of a from 2 sb ^, the exponent of a'. So, the result of the
J. 1 J-' J._L 1 i
division of a» by a» must be a"""", since a"» ""x a* a= a"». When
the denominators of the fractional exponents are difiiBrent, the subtrac-
tion can only be indicated, not performed, until they are reduced to
the same index. If the exponents of the same letter in the divisor
exceed that of the dividend/ the exponent of that letter in the quoticiit
■
i
EXTBACTION OV BOOTS. 229
tTe. So, also, if there are any letters in the divisor not
to the dividend, they mnst appear in the quotient, with their
re exponents taken with a contrary sign ; else, when the quo-
divisor are multiplied together, these letters would enter in
)duct. ,
RULE.
Divide the coefficient of the dividend hy that of the divisor , the restdt
he the coefficient of the quotient. Write after it all letters common
dividend and divisor, affected vnth exponents equal to the difference
their exponents in the dividend and divisor; and, also, all the
"s common to the dividend only, toith their primitive exponents;
all common to the divisor only, with their primitive exponents
ten with a contrary sign.
EXA1CPLE8.
'i, 1. Divide U^ch^ by 2a*cV. Ans. 2a*cd""'^,
if 2. Divide 12a ' ft • c V by 6a» 6«. Ans. 2a »' b'^ c df.
'^ 3. Divide 7x^y''zp by x»^»r"'. Ans. 7z p .
:ff
rKri !,„ iJJ A.. ^^...-..-*
■ 4. Kvide 7z y hj z y x . Ans. 72r'y~'a; .
5. Divide 24a*5*c* by 24z«a'^rV^. Ans. z^abc.
6. Divide 18x y z hj — 7 — j — r--
a;~*y"^~ -4n«. 1.
I "J. 11
7. Divide Sa''fh"<? by Ja-pfe"^<r- . Ans. 9b'<^d^.
8. Divide SOm^^ftM by ^mh'^x^. Ans. IBm^ftcV*
»
-1 * 1 ("-fn <i-H») (»-H>>
9. Divide 860e -/ ' ^ >• ^7 l^Oefg. Ans. 2c~ - / » ^ »• .
10. Divide 4a» c^ft^ by a-d:"*«"*. ^n8. ch^d^eK
20
230 FORMATION OF THE POWERS AND
RAISING TO POWERS QUANTITIES AFFECTED WITH FRAC-
TIONAL AND ENTIRE EXPONENTS. — MONOMIALS.
289. Since, to raise a quantity to a power is only to multiply it by
itself a certain number of times, the rule for the involution of quantities
affected with any exponents whatever is deduced directly from the rale
of multiplication.
RULE.
Raise the coefficient to the required power, and write after it aU the
literal factors affected with exponents equal to the product of their
primitive exponents hy the exponent of the power.
Thus, to raise a to the third power, is to take a ihree times as a
factor. Then, (a )• ss a X a X a == a = a'. The exponent
of the result is the product arising from multiplying the exponent of
the quantity by 3, the exponent of the power. So, (Mx* )»=(M)*x* ,
i_ i. L i.
because, (Mx")» = M X M X Mrc""*""""^" + &o. If the given quan-
tity be a fraction, it is raised to the required power by raising the
numerator and denominator, separately, to that power : because, the
power of a fraction is nothing more than the product of the fraction
multiplied by itself a certain number of times ; and a fraction is mul-
tiplied by itself by multiplying the numerators and denominators to-
gether separately.
EXAMPLES.
1 JL
1. Find the — ^^^ power of a". Ans. a ' .
2. Find the m** power of a ' . Ans. a^.
3. Find the «* power of a". ^n*. a"^.
BXTRAOTION OF BOOTS.
281
4. Find the — m»* power of a » .
6. Find the 4*»» power of 8aVJ*.
6. Find the 3* power of the 4** power of -=--.
2a
7. Find the g^* power of -=-.
8. Find the — 5m*^ power of aV.
Ant. a ^
Ans. Sla^cH'\
Afu, 12Sa'h-\
Ans. (2)'»a^J-^.
Am. a'-^^h"^
9. Find the — 5m*' power of the 8* power of aV.
Ans. a"*^&
10. Find the —12'* power of the —6** power of ^.
Ans. + a"6-".
11. Find the — 12<* power of the —7** power of — ^.
Ans. H-a'^ft"**.
12. Find the —11** power of the —7'* power of — •^.
Ans.
n-^.
13. Find the r<* power of the m** power of a".
Ans. a"*.
14. Find the — r** power of the m** power of a"*. Ans. a"".
15. Find the r** power of the — w** power of a"*. Ans. a^»
16. Find the — r** power of the — m** power of a'. An^. a"".
17. Find the 5'" power of
1 1
An*.
18. Find the 5** power of
ah^
19. Find the 6*' power of 2a*6*.
1 1
20. Find the 6* power of —2aH^.
21. Find the 3* power of 2a'6'.
Ans. -f
ah^
Ans. 64a'&.
Ans. 64a'5.
1
Ans. 8a5^.
282 FORMATION 09 THE PQWEBS AND
22. Find the 3* power of — 2a^6'. An$. —8a*'.
-Li. ' •.
23. Find the f^ power of Ma » 6 p. Ans. (M)*ali6p.
EXTRACTION OF ROOTS OF QUANTITIES AFFECTED WITH
ENTIRE OR FRACTIONAL EXPONENTS. — MONOMIALS.
290. To extract the m^ root of any quantity, a^, is to find a second
quantity which, when raised to the m^ power, will produce ihe given
quantity, a^. The extraction of a root is then just the reverse of raising
to a power, and, of course, the reverse process must be pursued to fiud
the root. The m* root of a^ must be am, because the m* power of a«
is equal to a^. To raise a quantity, which has a coefficient, to a power,
we first raise the coefficient to the indicated power, and write the new
coefficient before the literal factors as the coefficient of the power.
Hence, to extract the root of a quantity, which has a coefficient, we
must first extract the root of the coefficient, and write it before the root
of the literal factors. The m}^ power of Ma** is (M)"aP", therefore,
the m^^ root .of (M)"'aP°* must be Mqp. The m^ power of the fraction
a' a"' a"' a'
£- s= r--. Hence, the m* root of ^ir. = t;- That is, the m* root of
o
numerator and denominator must be taken separately.
RULE.
Extract the root of the coefficient, and write the result he/ore the Ut&'
ral factors affected with exponents equal to the quotient arising fiom
dividing their primitive exponents hy the index of the root If the co-
efficient of the given monomial is not a perfect power of the degree of
the root to he extracted, the operation is impossible. If the exponents
of ihe literal factors are not divisible by the index of the root, the lite-
ral factors win appear in the root with fractional exponents. The root
of fractions is extracted by extracting the root of the numerator and
denominator separately,
EXAMPLES.
1 2
1. Find the -** root of a '. Ans, a".
r
2. Find the r*** root of a'. '^'
BXTRACTIOM OF BOOTS. 288
8.
Find the m** root of a^.
1
Ans. aT.
4.
End the * root of a~.
r
Am. a-".
5.
Rnd the — ro"* root of a ' .
6.
Find the — r* root of a f" .
^««. a""**.
7.
Find the — »»* root of a'^ .
Am, a'r.
8.
1 ■
Find the ^* root of a~ ' .
m
9.
Find the 4* root of 81a»c»6'«.
AfiB. Sa*<^h*,
10.
Find the 7* root of 128a^6-'.
Ans. 2ah-\
11.
Find the ^ root of {2ya<b'^.
An$, 2aJ-".
12.
Find the —5m*'' root of a-^l"^.
Ans. a'ft*.
18.
Find the r* root of a"".
iliw. a*".
14.
Find the 4* root of l^Jn^.
Ans. 2a'Vj*.
15.
Find the 5** root of ^^\ .
c
16.
Find the 5* root of r-.
0*
17.
Find the 8* root of 8o6*.
Ans. 2ah^.
18.
Find the 6"* root of 64a%.
Ans. 2a*6*.
19.
Find the 8* root of — 8a6*.
Ans. ~2a*6*.
20.
1 «
Find the V^ root of (M)*a»6 p .
PROMISOUOUS EXAMPLES.
1. Bai8e2a ft" ?c^ to the 3* power. Ans. Sa'h'pc.
2. Extract the 3* root of Sa~^b F. Jlnt. 2a ^ITTc
20*
284 FORMATION OF THE POWERB AND
3. Multiply 2a-P6-"c— by 3aP6"c". -An*. 6
4. Divide 6 by 3aPZ>"c». Ans. 2a-«»i— <r*.
5. Divide 6 by 2ar^h-^d^. Ans. Za^hV.
6. Raise 2arlrc^ to the 6*' power. Ans, 64a"5c».
7. Extract the 6"* root of 64a*6c'. An$. 2aHK^.
8. Multiply (M)ia» by (M) « a » - Am. Ma.
9. Divide Ma by (M) » a » . -4»w. (M)Ba:.
10. Divide Ma by (M)i»aj. Ans, (M) ■ a ■.
4 «
11. Multiply Sx" V"* by ^. ' Ans. xy.
12. Divide xy by 8x"'y"i. ^iw. ^^.
o
18. Divide xy by —5^. Ans. Sx y *.
o
14. BAise ^ to the third power.
Ans. — 125x"y^jr*a"*6-*.
15. Raise Pa«5» to the r"" power. -4n«. (P)'a«'6*
16. Extract the r** root of (Pya^hK Ans. PaV.
17. Divide (2)V6« by (2)«a«5-. Ans. 2a« —ft"-.
18. Multiply 2a*— 5«— by (2)«a-6". ^iw. 8a'fi».
19. Multiply 2a»— 6*- by (2)?a*-5*-. ^n«. (2)*a»<H«)6«(H-).
20. Extract the »** root of (W)*a-J"^. ^n». (WycrV^.
21. Raise 3a ^6' to the 10** power. Ans. (3)»a'^6".
22. Extract the 10*' root of (8)% '6'. Ans. 3a V'.
28. Multiply (4)'a*6* by (4)' Ji^ Jn«. (4)»<«.
BXTBA0TIOI7 OF BOOTS.
24. Divide (4)V5 by (iyah^. Ans. (4ya*6*.
25. Divide (4)V6 by (4ya*6*. Ans, (4)«a^68.
26. Kaise — ah^ to the 3'* power. Ans. — aH^.
27. Kaise — arb^ to the 4** power. Ans, + a"6^.
ill
28. Raise 2j5«y«a« to the c** power. ilfw. (2)*ayV.
2 £ £
29. Extract the c<» root of (2yx^, Ans. 2a:«y«2«.
80. Find the r» root of ^^. Ans, 2a-?.
a"*
81. Baise 2a"' to the r^ power. ^n». (2)' a"*.
82. Raise a ••6* to the 10* power. Ans. aV.
33. Extract the 10* root of aW. Ans. a^b*.
84. Raise a»6« to the 2** power. ^iw. ab^*
85. Extract the square root of ab^'K ^ Ans, a-^b'\
86. Raise a"6" to the 6* power. Ans, a*»6'".
37. Extract the 6* root of a°"6'*". Ans. a"&-«
38. Raise a«»6"» to the 1000* power. Ans. a%\
89. Extract the 1000* root of aV. Ans. a*»6".
40. Multiply a*6* by a«6». . Ans. a'^b'^.
41. Divide a''6« by a*6*. -4n«. a»6".
42. Divide a'^b'^ by a»6». Ans, a»6*.
48. Multiply (10)«a'6» by (10)»a»6«. Ans, (lOya-^b-^.
44. Divide (10ya*6» by (10)«a»6«. ^tw. (10)"a-»6».
45. Divide (10)»a»6» by (10)»a«ft». Ans. (10)*a-»6-«.
46. Multiply (10)-'a*5« by (10)»a-*6-^. Ans. (10)«a»&«.
289 FORMATION OF THB POWSES AND
47. lUiw 2aB«c^ to the ^V P0^«- -^»«' (2)'a'6*c»
48. Extract the y^jtii root of (2)*a'6-»c*. ^n«. 2ayc*.
49. Baise (2yx\^:f to the y^a power. ^n<. (2)»x- V*«*-
50. Raise (2)»a»5» to the y^^a power. Am. 2a*"5^'.
It will he seen that decimal powers are smaller^ and decimal roots
greater, than the quantities themsdves.
CALCULUS OP RADICALS.
291. Any quantity with a radical sign is called a radical quantity,
or simply a radical. Thus, >/a, Vhy y/<?, are radical quantities, or
radicals.
The coefficient of a radical is the quantity prefixed to the radical
sign, and it indicates the numher of times, plus one, that the radical
has heen added to itself. Thus, 2^ a and m^a, indicate that y/a has
heen added to itself once and m — 1 times. When no coefficient is
written, unity is understood to be the coefficient. Thus, Va =s l>/a.
When the indicated^root can be exactly extracted, the radical is said to
be commensurable or rational. Thus, \/4 and v^ 8 are rational radicals.
When the indicated root cannot be exactly extracted, the radical is said
to be incommensurable or irrational. Thus, >/2 and v^5 are irrational
or incommensurable radicals.
A root has been defined to be a quantity, which, taken as a &etor
a certain number of times, will produce the given quantity. An «t)icii
root is a quantity, which, taken as a factor an cTjen number of times,
will produce the giy^ quantity. But no quantity taken an eren
number of times will produce a negative result. Hence the even loot
of a negative quantity is impossible. The indicated even roots of
n egativ e quant ities are called imaginary quaniiitei. Thus, y/ — 2,
V — 2, y/ — 2, are imaginary quantities. Roots that are not imaginaiy
are called real roots.
The term rational is in contradistinction to irrational.
The term real is in contradistinction to imaginaiy.
A quantity may be real and not rational -y but no quantity can be
rational or irrational and not be real. Thus, y/2 is real, but not
rational ; but y/i and %/ 2^ are both real.
XXTEAOTION OV BOOTS. 287
292. It has been shown that all ladicals may be changed into paren-
thetical expressions ; the numerator of the exponent of the parenthesis
denoting the power to which the quantity under the radical sign is
raised, and the denominator of the exponent of the parenthesis
denoting the index of the radical, or the degree of the root to be ex-
tracted. Thus, >/a^ may be changed into a^^ and ^o* may be
changed into a'.
293. A simple radical is one, which, when changed into an equiva-
lent parenthetical expression, has unity for the numerator of the
exponent of the parenthesis. Thus ^a, f/a^ &o., are simple radicals.
A complex radical is one which has the quantity under the sign raised
to some power different from unity. Thus, 'Ja% Va", are complex
radicals, the equivalent parenthetical expressions (a)^, and (a) ", having
numerators different from unity.
294. Radicals are said to be similar when they have the same index
and the same quantity under the sign. Thus, 2y/a and Sn/o are
similar radicals. The ^a and -^a are not similar, because, though
the quantities under the radical signs are the same, the indices of the
radicals are different. The >/a and >/6 are dissimilar, because the
quantities under the signs are different.
295. Similar powers of the same quantity can be added by adding
their coefficients. Thus, 2a' -f 3a' = 5a'; because, since the literal
factors are the same, they may be represented by the same letter, x.
Then, 2a" + 8a' = 2a: -f 3x = 5x, and replacing x by its value a", we
have the sum of the two quantities equal to 5a'. For a like reason,
similar powers of the same quantity may be subtracted from each other,
by taking the difference between their coefficients and uniting this
difference as the coefficient of the common quantity. Thus, 5a*—
2a* = 3a'.
296. In the same way, similar radicals may be added or subtracted,
2ya-f >/a = 3%/a. For, make V a 3= a; ; then 2Va-f ^/as=s2a;+
a; =b 3x bs Sy/a. So, also, 2>/a — ^a ^ Va.
297. The calculus of radicals shows how radicals may be operated
upon algebraically — added, subtracted, multiplied, fto., ftc.
These algebraic operations must be in accordance with certain prin-
ciples, which both modify and facilitate them. It would require an
extended treatise to embrace all the principles of the calculus of
radicals. A few of the most important only are given in this work.
288 FOBMATION OF THE POWERS AND
First Principle.
298. Any parentlietical expression^ composed of several &ctois, may
be decomposed into as many new parenthetical expressions as there are
p p p p
faotoiB. Thus, (a6c)q = (a)q (i)q (c)q .
For by the rules for raising a monomial affected with any exponent,
fractional or entire, to a power, each fiictor has to be separately raised
p p p p
to the indicated power. Hence, (a6c)q = (a)q (h)T (c)q" . And, in
p p p p
like manner, (a^ft'c^q = (a"X (6°)^* (c')q .
299. This principle has an important application in the case of simple
radicals. For (ahcd)^ = Va2>c(£, isalso equal to (a)» (2>)n (c)n (dy
or to y a". yT*. y 7;^^. Hence, J^ahcd = ^'a^JyTy'd. That
is, the n** root of the product of any number o/ factors is equal to the
product of the n^ roots of these factors.
800. This property of radicals is used for two distinct purposes.
1st. To simplify radicals. Let it be required to simplify or reduce
^/ 75a'c. We see that the quantity under the sign 75a% can be de-
composed into two factors, 25a' and 3c, and that the first is a perfect
power of the degree of the root to be extracted. Then, V 75a^c =
\^25a' X 3c, which is also, by the property just demonstrated, =
^/2ba^y/Bc = 6a ^/3c^ In like manner, ySa'c = ^8^ X VT=
2a Vc. Then, to simplify or reduce a radical, we have the following
RULE.
Decompose the quantity under the sign into ttco factors, one of which
shall he a perfect power of the degree of the root to he extracted. Ex-
irojct. the root of the perfect power , and write this root hefore the incom-
mensurable factor.
801. 2d. The property, that the n^ root of the product of any number
of factors is equal to the product of the n^ roots of those factors, is also
used to make radicals simikr which appear dissimilar. Then, after
they have been made similar, they may be operated upon algebraically,
added, subtracted, &c., like simple algebraic expressions.
Thus, i/4a'c and mVc" appear to be dissimikr. But v^4^ =
V4a' . c = \^4a" y/c = 2ay/c. Hence, if it be required to add >^4a*c
and m >/ c, we may represent Vchjx, and the sum of the two quantities
XXTBAOTION OF BOOTS.* 289
will be 2ax + mx or (2a + m) x. Now replace x bj its value >/ c, and we
have ^^a^c -^ my/c =^ (2a + fn)n/c. In like manner, 26^ 8aV
and — 2ac y c^ appear to be dissimilar. But, 26y8aV =s
2fty8aVx"? = 2&y 8aV X_y^ = 4ac5y7, and — 2ac V^B" =
— 2acy 6»(c«) = — 2ac ^J^ V^ ^ — 2ach y?. Hence, 26 V 8aV —
2ac^c?6' s 2a€h^<^. To operate upon dissimilar radicals, we have
then the following
RULE.
First make the radicals similar, then represent the common radical
hy X, or conceive it to he so represented, and operate upon the new radi-
cal expressions according to the rules for simple algebraic expressions.
The following examples will afford illustrations of the two applica-
tions of the principle, that ^abc, &c. = v^a . yb ^c, &c. For the
sake of simplicity, we will assume the own signs of all the radicals to
be positive.
EXAMPLES.
1. Simplify the expression V16aV6'. Ans. 4tacVy/cb.
2. Simplify the expression yil^'^z?. Ans. 3xi^s^y/z.
3. Simplify the expression V 50a'6V. Ans. 5ab^c^2abc,
4. Sunplify the expression y24a*6V. Ans. 2a6Vy3a«.
5. Add the radicals V16a'6V and + 2al^cVa.
Ans. 6al^ey/a.
6. Add the radicals v'60a'"6^ and + 6a»6»V26-^.
Ans. 10a"ft«V2^
7. Add ihe radicals 4 Va"'6-'<rP and + 3a"'6*c^^a-*^5-*c-"p.
Ans. 7 Vcr'6- ■<?-*•.
8. Add the radicals 4a'^ Vc? and 8c ^aH*c.
Ans. 12a'5•c^/c.
9. Reduce \/27 + '/i2 — -/75 to one sum.
Ans. 0^3, or 0.
240 FORMATION OF THE POWEBS AND
10. Reduce V50 + V8+ y/M'+ >/18 + %/162 to one 8iiiii.__
An$. 23^/2.
11. Reduce 2 ^c + 4 y"? + 2a ^"7" to one Bum. ^
^n*. (2 + 4c + 2ac^^c;
12. Reduce a>/20 4- hy/6 — ci/66* to one sum. ^
Ans. (2a + & — ch)^b.
13. Reduce a5x/72 + cV8 + mn y/lSc* to one sum.
^•i». f6a6 + 2c+8innc)>/2.
14. Reduce 5^8 and V 8a*c" to one sum. -4n«. (5+ac^y8-
15. Reduce 12^ a^b'^d and m^cT. Ans. (i2ab^ +fnd)Vd.
16. Reduce a^6*+*c*+*n and my n'+'ftV.
Ans. (ahc + mn) -V 6*<rSi.
17. Reduce >/ 48 + 6a ^/T2 + 66 ViS — 246 >/ 12. _
-irw. (2 + 5a — 126) >/ 12.
18. Reduce 4av'c' + 5acV<^"""0ac>/c to one sum. Ans. 0.
19. Reduce 126 V a^^c*" + 5m V a to one sum.
Ans. (126ac" + 5m) Vol
20. Reduce S^a^t^b^ and — 2abcy/abc to one sum.
^n<. abcn/aic.
The whole difficulty in solving these examples consists in finding the
incommensurable fiictor common to all the radicals.
Second Principle.
302. The n^ root of the quotient of two quantities is equal to the
quotient of their n* roots. That is, \ / -7- = —^=^
For, to raise a fraction to any power, we raise the numerator and
denominator, separately, to the required power. Hence, to extract any
root, as the n*'', the root of each term of the fraction must be eztncted
separately.
This principle is used in extracting the roots of fractions.
IXTBAOTIOM or EOOTS. 241
Third Principle.
303. The mn^ root of any quantity is equal to the m^ root of the
n** root of tibat quantity. That ia, 7/0"= 4 / jya,
A quantity, a, may be raised to the sixth power by first squaring a,
and then cubing the result. Hence, the sixth root of a' might be
extracted by first extracting the square root, and then extracting the
cube root of the result. So, a may be raised to the mn* power by
first raising it to the n"* power^ and then raising the result to the nC^
power. Hence, the mn^ root can plainly be extracted by taking first
the n% and then the m^ root, in succession.
304. This principle is of great importance, and extensive applica-
tion. It is used to extract high roots in succession, whenever their
indices can be decomposed into fiiotors.
Thus, V^5B5^= </ v^"55B?5^ = v^"I55^ = 4a«i».
411 even roots, and roots that a^ multiples of 3, can be extracted in
tfooeefision. The sixth root is equal to the cube root of the square root.
The eighth root is equal to the fourth root of the square root, or it <»ui
be found by extracting the square root three times. But the fifth root,
seyeoth root, eleventh root, &c., cannot be extracted successively.
WLen the principle can be applied; we begin by extracting the low-
est root first. The index of the radical must be decomposed into its
prime fiEtctors, and the root corresponding to the lowest factor ought to
be ^unst extracted. It frequently happens that the factors are equal.
EXAMPLES.
\. Extract the eighth root of 256a%^
2. Extract the sixth root of 729a'®6".
y729a"M* = s^ ^l^a^ = V27a»J* = 8a»6*.
8. Bequired V6561a>«6«. An$. 9aW.
4. Bequired \/WScl¥. Am. 4a*i*
21 Q
242 FORMATION Of THB POWEES AND
6. Eequired ^1095^. Am. 4a*ft*
6. Kequired ^^665360^6^. An». 4a*6'.
7. Required ^6561a»»6*«. Aru, 3a»6».
8. Eequired ^a^h^. Am, al^.
9. Required y(27)»aH"-. Ans. SM\
In this example, extract the s^ root first.
Fourth Principle.
305. The coefficient of a radical may be passed under the radical
sign by raising it to a power indicated by the index of the radical.
That is, P V^ = yP^. For, P= yP^; henc e, P Va = yP^ V7;
b ut, by the first principle, 7P» 7a = 7P-a. Therefore, Pya =
7P-a.
306. The fourth principle is, obviously, just the converse of the
first principle. The latter enables us to pass a factor outside of the
radical ; the former, to place a fiictor under the radical. The fourth
principle is frequently used in the differential calculus. It has two
applications in Algebra; Ist. The approximate value of incommen-
surable roots can sometimes be found more exactly by means of this
principle. Thus, | y/T= V(|)«.7= y/^ = | : true to within less
than I . A nearer approximation to the value of the expression has
been found in this instance by passing the coefficient under the radical.
But, for such expressions as |v% |\^5, nothing is gained by passing
the coefficient under the radical.
2d. The fourth principle may be used instead of the first, or in con-
nection with the first, in making radicals similar which appear dissimi-
lar. Thus, 9a>/6, and ^/Sla'h are dissimilar, until either 9a is
passed under the radical, or 81a' is passed without. Sometimes it is
difficult to employ the first principle to make radicals similar, because
it is not easy to perceive the incommensurable fstctor common to all the
radicals. But there is no difficulty in employing the fourth principle.
Pass all the coefficients under their respective radical signs, and then
EXTRACTION OF BOOTS. 243
decompoee the numerical and literal faotors into their prime factors.
If the radicals can be made similar, common factors will become appa-
rent after the decomposition into prime factors. Let it be required to
ascertain whether 2ay/ah^, and b^a*b* are similar. The equivalent
expressions are y/Sa'^b^, and \/a*b^. The two radicals have, then, a
common factor, a*b^y under the sign. This common &ctor can be de-
composed into two others, one of which is a perfect power of the degree
of the root to be extracted, and the other incommensurable. The in-
commensurable factor may be made a new radical, and the given ex-
pressions being decomposed into &ctors, one of which is this incom-
mensurable factor, will be similar. We have, in accordance with this
rule, 2ay^= j/'Sgfc^ = V^^&j/^^ = 2ab^W, and 6{^d^
Take, as a second example, ly/a:^^, and a;^/2401x y; by passing 7
and X \inder the radical signs, we have 4^2401x'^, and y2401x^.
Hence, V2401x^ is a common factor, and the given expressions reduce
to Ixi/xt^ and *lxi/x^j and are not similar, since they have no com-
mon incommensurable factor. We have, then, a simple process for
ascertaining whether radical expressions can be made similar.
RULE.
Pais the eoefficienU of the different radicals under their retpeetive
tigm, and then examine whether there w a common incommensurable
/actor, 1/sOf the radicals may be made similar by taking the income
mensurable /actor as a new radical. Write be/ore ity /or new coejffi-
dents f the roots o/ the eomm^nsur<ible /actors 0/ the respective eoc^es-
sums, and we wiU have a series o/neiw radical eapressionSf aU similar.
If there is no common incommensurable /actor, the radical expressions
are not similar.
EXAMPLES.
1. Add together 2a V&? and dc5^x/5i^.
Then , 2a Vi?= V4fl?6? = y/ltcMy/bc and ScI^Vb?^ >/W? =
^Wi? Vbc. Hence, V^ is a common incommensurable factor, and
the sum is (2ac -h Sbc) V be.
2. Add 4a5y >/aV and acz>/ab*c together.
Ans. (4a5 y -f Vz) qa^Joc.
244 FORMATION OF THB POWERS AND
8. Add together 6>/640a» and 6a ^/ 240a.
ftV"54 0?= V 3.5.2«.3Vy and 6a v/ 240a = >/ 36.3.5.4V.
Hence, >/3.5.a, or -/ 15a, is a common incommensurable factor, and
ihe sum of the two expressions is (fiah + 24a) >/ 15a.
4. Add together v^'SO and V810.
^90 = V3«.5.2 and %^ 810 = >/3\2.5. Hence, v^2T5 is tlie
common incommensurable factor, and the sum is (3 + 9) %/ 10.
5. Add together a V 216 and 6 %/ 2400. An9. (6a + 20i) V 6.
6. From a v/'lOO subtract ft ^"243. Am. (lOa— 9&)v^3;
7. From a V 33750 subtract 6 5^ 10000.
I
Ans, (3a— 2i) -3^1250.
8. Add the three expressions, ha^ax, mx V9«*j «<> V^'
Ans, (ba + 3ma + n)y/ax.
9. Add the three expressions, ha^o^y «wj* V"27a* + na^^x.
Ans. (ba + 3ma + n')yax.
10, Add the three expressions, hai/cSc + «i.«* y 81a* + marj/x.
Asu, (ha + Bma + M)yax.
JY/llA Pnncipfe.
807. Any &ctor may be passed without a psrenthedoal ezpresakm,
by multiplying its exponent by the exponent of the parenthesis.
p 5P p
Thus, (a*5c)l= a «» (6c)q.
For, (a'^bc)'^^ a'^b'^c^, since each term hafl to be rused to the ^
power. And, by separating into factors, we haye ai X b^c^, which is
also equal to a^ (6c)«.
If there is more than one term within the parenthesis, any term or
a factor of any term may be passed out, by dividing all the terms by
ihe expression to be passed out, and then writing that expression out-
side of the pareathesisy with its primitive exponent multiplied by ihe
XXTBAOTIOH OF E00T8. 245
exponent of the parenthesis. Let it be required to pass a* without the
parenthesis (a"6 + c) <. This may be written a"(^ + -j I p = (aJ^d^
= a < (<Qr In which ci = 6+ — o(r6 + a""c. HencO; (a"6 + c)<
■p p
It matters not how many terms there may be within the parenthesis ;
they may be all represented by a single letter, after they have been
divided by the factor to be passed without. The foregoing demonstra-
tion will then be applicable.
EXAMPLES.
1. Pass X without the parenthesis (scy'i*) *.
Am. cT^Q^Tff^.
2. Pass a;"* without the parenthesis (a;~*j/"2')'.
Ant. »"»'(yV)».
3. Pass x~* without the parenthesis (a5~*a5)"*.
AnM. «(a6)"*.
4. Pass 4 without the parenthesis (4a%»)*- ^««- SCa'ft')*.
£ IS. 1
5. PteB a^ without the parenthesis (a'j/^af)'. Ant. x(3f^^y.
6. Pass 8 without the paienthesis (8a&)^. Am. 2 (ab)^.
7. Pass 16 without the parenthesis (16a&) . Ans. 2(ahy.
(4a + ft)^ / h\i
8. Pass 4 without the parenthesis « . Ans. \<» + j/ •
9. Pass 5 without the parenthesis (a + bby.
Ans. 125 (I + iV
10. Pass 2 without the parenthesis (2a + 45/.
ilfw. 16 (a +2hy.
21*
246 FORMATION OF THE POWE&S AND
^ ^ . (4 (mo; + na:«) + (»» + 2na;)* )*
11. Pass 4 without the parenthesis s^ ■
Ans,
(mx + nx" + J (m + 2nx)* )5
m*
12. Pass the term 4a without the parenthesis (4a + & + c) .
^„,. 2„Ki+^^^^^y^)*
13 Pass the term 8a^ without the parenthesis (8a' + & + c) .
. « ^1 . a"'(^+c)\*
^n«. 2aVl + — ^ — ^) -
14. Pass the term 5a' without the parenthesis (5a' + b + cy.
An.. (sMl + ?^')'.
Any number of terms may be passed without^ by representing them
by a single letter, and then replacing that letter by its value after the
transfer has been made.
15. Pass (a + li) without the parenthesis (a + 6 -f- c)*. Let a+ h
= X. Then, (a + 6 + c)* = (a; -h c)* = x*(l + x-'c)^ = (a+ 6)*
|l + (a + 6)-'c|*.
16. Pass (a+h -\-c) without the parenthesb (a + b + c + dy,
Ans. (a +i + c)* I l4-(a + i + c)-'d j*
17. Pass (a'\-h +c) without the parenthesis (a + ft + c)*.
Ans. (a+ 5 + c)« (1)' or (a + b + cf.
1
18. Pass or* without the parenthesis (a"" + ft)**.
Ans. a p(l 4- a'ft)^.
The last example shows that any factor affected with a negative ex-
ponent may be made to appear with a positive exponent in the other
terms of the parenthesis. This transformation is used in the differential
calculus when it is desired to change a negative into a positive exponent.
It is evident that the fifth principle is only the more extended appli-
cation of the first principle.
BXTBAOTION Of BOOTS. 247
Sixth Princi'ple.
808. Any factor may be passed witUn a parenthesis^ by multiplying
its exponent by the reciprocal of tbe parenthesis.
Because, to pass it out again, we must multiply its new exponent by
the exponent of the parenthesis, and when its new exponent has been
formed as directed, the factor, after it has been passed out again^ will
p ■? £
be affected with its primitive exponent ThuS; a"(5)^ =s(ap6)*i.
Because, when a^ is passed out again, the expression will become
■q p £ p
EXAMPLES.
1. Pass the coefficient 2 within the parenthesis (a + &) •
Afii. I (2)«a + (2)«6 I*.
2. Pass the coefficient 2 within the parenthesis (a + h) .
Am. (8a + 85) .
i 1.
3. Pass a* within the parenthesis (a» + ^)'* •
Ans. (ap p + apft)^.
_"jq p.
4. Pass a" within the parenthesis (a p + 6)«i .
■? p.
Ans. (1 + a"? 6)7.
5. Pass a" within the parenthesis (a^ + 5)
-4n». (a"^ + fta"*") .
6. Clear 8(|a + J6)* of its coefficient. Am. {^ + 6^^.
7. Clear 64^ .^^^ + "t^ of its coefficient.
An.. (I + Z2cf.
8. Clear 27(2a + ]fi)^ of its coefficient Ans. (18a + 6)f
248 FOBMAVlOtf OF tnt P0WSB8 AND
9. Clear 32 (2a + ib)^ of its ooeffioient. Ans. (8a + by.
10 Clear the parenthetical expression 125(5^a + gtV of its
^®*^^*- Ans. (25a + 1256)f
11. Clear the parenthetical expression a~(l -f- a"*)« of its coefficient.
r
jln«. (a 4- 1)^-
It is evident that the sixth principle is only the more extended apr
plication of the fourth principle.
Seventh Principle.
309 . The denominator of the exponent of a parenthetical expression may
be multiplied by any quantity, provided we raise the quantity within
p
the parenthesis to a power denoted by the multiplier. Thus, (a)fl =
£ P JL _L ■?£ -H. i -£.
(a")"«». For (a)T = a q , and (a^)"^ = a""* = a <» . Hence, (a) «» = (O"'-
If there is more than one term within the parenthesis, their algebraic
sum may be represented by a single letter, and the foregoing demon-
stration is, therefore, applicable to all kinds of parenthetical expres-
sions.
310. The seventh principle has two applications. 1st. It is used to
cause complex radicals (or parenthetical expressions with the nume-
rators of their exponents different from unity), to be affected with ex-
ponents having a common denominator.
Thus, (a + by and (a + &)', can be changed in accordance with the
principle into the equivalent expressions ( (a -f- 6)*)' and ( (a + &)■)•.
We see that a second parenthesis has been written within each pa-
renthesis ; and we see, also, that the exponent of the first new paren-
thesis is the quotient arising from the division of 6, the least common
multiple of the denominators of the exponents of the given parentheses,
by the denominator of the exponent of the first given parenthesis.
The exponent of the second new parenthesis is the quotient arising
from dividing the same least common multiple, 6, by the denominator
of the exponent of the second given parenthesis. In like manner.
XXTEAOTIOBI OF BOOTB. 249
(a + by and (a + b)* may be changed into ((ft + *)*) and
((« + &)») A
Hence^ to oause parenthetical ezpressionB to be affected with expo-
nents having a common denominator; we have the following
BULI.
Take the leaH common mtdtipte of aU the denominator of the ea^-
ponents of the parenthezesj and divide thai multiple by the denominator
of the exponent of every parenthens. The several quotients will indi-
cate thepovjer to which the quantity within their respective pareiUheaes
must be raised.
EXAMPLES.
P P.
1. Change (a + &)" and (a 4- by^y into equivalent parenthetical ex-
pressions; the denominators of whose exponents shall be the same.
p_ jp_
Ans. ((a + 6)»)«and(a + 6)-».
JL 2-
2. Change (a -h by^ and (a + 6)"^; into equivalent expressions.
. p p.
Ans. {(a + bY)^ and ( (a + 5)- ) ■«»,
311. 2d. But the most important application of the seventh prin-
ciple; is in reducing simple radicals to a common index.
ThuS; since a = a*, or va = i/c?, and a sbb a®, or ya ^ yo",
it is plain that >/a and y/a, can be reduced to a common index. So;
v^a and v^a, may be changed into the equivalent radicals *^a" and
%^a". In each instance; the power to which the quantity under the
radical is raised is indicated by the quotient arising from dividing the
least common multiple of the indices of all the radicals by the index
of the radical under consideration.
BULB.
Form the least common multiple of the indices of aU the radicals.
Raise the quantity under each radical to a power indicated by the qwh
iient arising from dividing the least common multiple by the index
of the radical under consideration.
250 FORMATION OF THE P0WSB8 AND
Beduce ^a, i/a, and ^a, to same index. The least common mul-
tiple is 12, and the three quotients ^ = 4, \;* = 8, and ^^ = 2.
Then the equivalent radicals are 9^ a*) y/l?, and Jy^'.
EXAMPLES.
1. Reduce ^a, Vo^ y/a, and {^a, to same index.
AfiB. V*^ yV, ^"oS and ^.
2. Beduce Va, (/a, y/a, ^i/a, and ^o] to same index.
An$. ^^, y?«, "y^, y^, and y?:
3. Reduce ^a, V«j v'o, V a, and v^oj to same index.
Am. •"•Jp'o", "•To", "Vo", "Vo^ and "^o^
4. Beduce </a7 V"a7 >/a^ Va, and y a~to same index.
5. Beduce ^, ^/"a, ^o, ^o", l^o^ Vo", and Vo^ to same index.
An$, y"^, V"^, -y"^ V^ V^y^and ^IT
6. Beduce Vo^ "ya7 Vo^ v^"a^ ^a7and "*y^to same index.
Am. 'Vo™^ ■Vo^ -y"a", "V a"', ^y a^ and -^ a%
7. Beduce ^a, ^a, Va, ^a, ^a, ^a, Va, and ^a, to same
index.
^n«. «»/"^ "s^y*?*^ *^a«^ «^^, "'Va^, «*5^^ "v^c?^
and **^'^.
It will be seen that simple radicals are changed into complex by the
operation of reduction to a common index.
Eighth PHnciple.
312. Any factor of the index of a complex radical may be sup-
pressed, provided, the same factor is suppressed in the exponent of
the_power to which the quantity under the sign is raised. That is,
For, "Va- = (a)"» = (a)» = yi.
SXTBAOTION OF BOOTS. 261
This principle is just the converse of the hist, and reverses the results
of the last. It is nsed to simplify complex radicab, and frequently
reduces them to simple radicals.
EXAMPLES.
1. Simplify Via + 6)*, V(a + h)\ Via-k- hf and !y (a + h)\
Am. y/a +6, Va •\-hy i/0"\- b, and y/a + h.
2. Simplify </(a + by, ^/{a + 6)^ !J/(a + h)\ and ^(a + by.
An$. y/ia^bfj (a + 6)", y/a + b, and V(a + ^y.
8. Simplify ^/(a+6)^ V(a+^)', v^ («+*)" V(a+*)*; </(a+ft)*
^n». (a + &)», y(a + 6), V(a + 6/, (a + 6), (a + 6)*
4. Reduce Vo^J V < y o?; and V^
-Afw. ya, Va, ya, and Va^
6. Reduce y^, 5^"^, V^, V^ and y^
jln«. V<z? v^^'i V^^ y/^} andv^a.
6. Reduce "Vo^, ^C/o^, '"Va", "Vo^, "Vo^.
^»w. ^a, Va, "^a, Va, and -J^a.
7. Reduce ?j/a», 5^"?, ?J/"a^, ^V, ?/a^ and ?/Z
^n«. ^T, 5^0", y o^ !}^ a, V a", and Va.
8. Reduce "V^c?^ "^0% "Va", 7/ a", "Va"", and "V a'.
J.n«. y a, "ya, Vo^ ^ a, v^a, and "Va.
9. Reduce ««5^a«», «?^a»«, »ya« "^ a««, "^a**, »»?^«*», »^a>»,
Ans. •J a, y/Qj y/a, y/a, y/a, ^a, Vcr, and Va*
313. There are two consequences of the last two principles^ of con-
siderable importance. 1st. Whenever it is required to extract a root
of a complex radical^ which is a multiple of the exponent of the quan-
tity under the sign, the extraction can be indicated by suppressing the
exponent under the sign^ and multiplying the index of the radical by
the quotient; arising from dividing the index of the required root by
252 FOKMATION OV THE P0WXB8 AND
the exponent under the sign. Thns, let it be required to take the
dxth fx>ot of •^/^, the resnlt will be ^a. ¥oTf the sixth root of V^
I 1 JL 1
is eqnal to (ar)^ = a^ =za* ^ ifa. In like manner, the pn^ root
of y a" = 'y a. For the pn**" root of V a» = (a^y^ = op" s= a»* =
'yfa. These results have plainly been formed in accordance with the
rule.
314« 2d. Whenever it is required to raise a simple radical to a power
which is a factor of the index of the radical, the operation can be per-
formed by dividing the index of the radical by the exponent of the
power, and writing the quotient as the index of the radical instead
of the old index. Thus, let it be required to raise Va to the square
power. The result will be Va; for the second power of the sixth root
12 1 __
of o is (a*)" « a* = a' =8 l/a. So, likewise, the n* power of Va
i_ 1 J.
= y/a. For the n*^ power of ^jCa at (cfl^y =s op" = o» =^a.
The following are applications of the consequences.
EXAMPLES.
1. Eequired 4** root of ^a\ Am. l/a, or a.
2. Required 6** root of v'a*- Ana. i/a.
8. Bequired 7^' root of v^a'^ Ans. a.
4. Required 4** root of V«"« -^««. (/a-
5. Required pm** root of v^^o^. Ans. ^"a,
6. Required 12«» root of \/o?. Am. J/oT
?• Required 4*>» power of Va^ -4*w. «•
8. Required m*" power of v/«^ Am. a.
9. Required win** power of ?j7a. Ant. \/a^ or «■.
10. Required m*** power of "v^a. Ans. y/a,
11. Required 8* power of ^li. Ant. i/Z
BZTBAOTION OP BOOTS. 258
12. Beqoired 12*^ power of \/a. Am. i/a, or a*.
IB. Bequized 6^ power of %/a. Ans. i/a,
14. Beqvired 5<^ power of i{/a. A'm. l/a.
15. Required jpm** power of '^a. f An9. y/a.
TO MAKB SURDS RATIONAL BY MULTIPLICATION.
CASE I.
Monomial Surdi.
815. Suppose ihe giren surd is y/T; tiiis is eqtuyftlentto ft*". Now,
1
it is required to multiply 5 b by fuch a quantity as will make it rational.
Since the surd will be rational wben the numerator of the iractional
exponent is exactly divisible by the denominator; and; since^ in multi*
plication^ we add the exponents of the same literal factors^ the multiplier
of 5' must be h, affected with such an exponent, that, when added to
— ^ the sum of tha two exponents will be a whole number. Call x the
unknown exponent of the multiplier, then x •\ = 1, ora;=:l
ss . Hence, the multiplier is £ " , and we see that ft* x 5 >
s= 5, a mtional product. Had we placed x H » 2, and found the
multiplier under this hypothesis, ihe prodnet woold have beea 5^.
But, when it is required to make the surd rational, and of the first
degree, the sum of the primitive exponent, and the unknown exponent,
must be placed equal to unify.
Let it be xequired to find a multipUer which will make y* rational.
Then, fc + f = 1, or x =s I. Hence, the multiplier is y*^ and we see
ihat y* . y* = y is a lational product.
22
254 FORMATION OF THE POWERS AND
RULE.
Place the primitive exponent, plus x, egwd to unity ; find the value
of T from this eqitation. The value 90 found toiU he the exponent of
the multiplier. The multiplier itself must be the given monomial, ex-
elusive of its exponent, raised to a power indicated by the value ofx,
f
EXAMPLES.
1. Find a multiplier which will make x"^ rational.
Ans, X ■ .
2. Find a multiplier that will make y^ rational. Ans. y^,
3. Find a multiplier that will make y > rational. Ans. y ■ •
4. Find a multiplier that will make y^-H" rational.
Ans. y •+■ .
CoroUary.
816. The principles demontrated in Case I. enable us to find the
approximate value of fractions whose denominators are monomial surds.
BULE.
Multiply both term^ of the fraction by such a quantity as v>m make
the denominator rational. Approximate as near as may be desired
to the true value of the monomial surd in the numerator, and then re-
duce ihefracHon to its lowest terms.
EXAMPLES.
5
1. Itequired the approximate value of — = to within *01.
>/10
Ans. 1*68.
Pnr ^ _6%^10 6(816) 16-80 _^
EXTRACTION OF BOOTS. 255
2. Bequired the appiozimaie value of —=: to witbin *001.
Am. 1-782.
4
3. Bequired ihe approzimato value of — =: to witliin *01.
Ant, 2-52.
4. Bequired the approximate value of — = to witUn '001.
^ viooo
Ans. 8 162.
5. Bequired the approximate value of —=z to within *01
Ans, 4*65.
6. Bequired the approximate value of —^= to within -0001.
^ ^^ V^IOOO
Ans, -189738; nearly.
7. Bequired the approximate value of — ==. to within -1.
y400
Ans. 54-2.
8. Bequired the approximate value of — = to within -01.
Ans. 2-32; nearly.
64
9. Bequired the approximate ^'alue of ■ to within -000001.
^ v'320
Ans. 3-577781, nearly.
g
10. Bequired the approximate value of — = to within -001.
V32
Ans. 2-519.
15
11. Bequired the approximate value of — = to within -001.
V 1^
Ans. 3-873; nearly.
30
12. Bequired the approximate value of —= to within *001.
Ans, 3-873; nearly.
13. Knd the approximate value of -^ to within -01.
Ans. 504; nearly.
266 FORMATION OF THE POWERS AND
o
14. Find the approximate value of —= to witbln '01.
Am, -89.
18
15. Find the approximate value of — =
^ yi8' An$. 6-87.
CASE n.
317. To find a multiplier that will make rational an expression^ con-
sisting of a monomial surd; connected with rational terms^ or consisting
of two monomial surds.
Let it be required to make rational ^/p + ^Jq by multiplication.
From the principle demonstrated in Case Ly it is plain that >/^ can
only be made rational by multiplying it by •//>] and \/^^u} only be
made rational by multiplying it by -^q. But, unless %/p and ^/q^ in
the multiplier^ are connected by the sign minus, there will be two terms
in the product remaining irrational and unreduced. Hence, the mul-
tiplier must have the minus sign between its terip^.
Thus, VF+v^"^
p+ y/pq
— y/pq—i
If the givenjazpresoon laVp^^ ^fq^ the multiplier must be, for a
like reason, ^-p + ^q.
If the given expression contain but one monomial surd, and is of the
form p+ %/ J, it maybe written /p^ + y/q, and reduced, as before,
byjnultiplying by i/jf — Vq^ So, p — y/qmnj be_written y/p —
Vq, and may be made rational by multiplying it by \/p^+ y/q'.
It mattera not how many terms may be under the radical, or how
many rational terms may be outside of the radical, the foregoing proces-
ses wiU still be applicable; because the sum of the quantities under the
sign may be represented by a single letter, and the sum of the rational
terms outside of the radi cal may be^ represented by a single letter.
Thus,_a + 6+ ^m-^n^p+ ^/y « v^p"+ v^*^ Multiply now
by y/]^ — y/q, and replace p and q by their values. The result will
be(a-f- 6)« — (m — »).
XXTRACTION OF AOOTB. 257
Let it be required to make Vp -\' Vq rational by multiplioatioii.
Tbe operation is as follows :
yp + V7 _
Vf+ V^ Vpq
+ Vpit + q
— Vp'q—^p^
p -|- ^ ss Product.
The moltiplier, to make y/p rational, most be l/j^ ; and the multi-
plier^ to make ^q rational; must be %/^. But, after multiplication
by Vp* + y/^y there remained two uncancelled surds, Vp\ and
V ^Pj and these could only be cancelled by multiplying the given ex-
pression by — Vpq'
To make rational l/p — Vq^ the multiplier must be V]f ■\- V(f
+ Vp^' ^^® multiplier, then, in every case, is the sum of the cube
roots of the squares of the quantities diminished or augmented by the
cube root of the product of the two quantities, according as the sign
between the surds is plus or minus.
Let it be required to make Vp ■\- i/q rational. The operation is
as follows :
Vp + y/q = Given surd,
Vp— V^ — Vpq + Vp? = Multiplier,
P +V^
-Vp<f-q
Vp'q—Vp'(f
+ Vf(f + vj?
p — 9 = Product.
Or the operation may be performed in this manner :
Vp + Vq == Given surd,
Vp — Vq-= First multiplier,
\/p— V(f= VTj- ^/l= Fi«* Product,
Vp + Vq= Second multiplier,
p — q= Second product.
Any two monomial surds, whose common index is some power of 2,
may be reduced in the same manner. And, since each multiplication
22* R
258 FORMATION OF THE POWSRS AND
giveA a prodnet containing two surds^ with a common index ane*half be
great as the common index previous to multiplication^ it is evident
that the number of multiplications will be indicated by the exponent
of the power of 2 in the primitive index, common to the surds given to
be reduced.
Thus, %/p + ^2 ^^ ^ reduced by three multiplications, since
8 s 2^. And x/p + \/q can be reduced by four multiplications,
Hittce 16 =r 2^ ' _ __ ^ _
The reduction of the 5/p + Vy> and l^p + l^q, is as follows :
Vp + ^9 = Given surd,
^p — ^q = First multiplier,
^~p^ %/^ s X/p — V? = First product,
X/'P + s/q = Second multiplier,
X/l^ — t/j*= \/p — y/q ^ Second product,
V'jP + y/q = Third multiplier,
p — q^^ Third product
S<P + v^T= Given surd,
v^P — y/q = Fij^* multiplier,
5/p»— !5/7= ^F— W== Fi"* product,
Vp + ^S' = Second multiplier,
%/^ — ^2*= 4/l> — VS' == Second product,
X/p + V? = Third multiplier,
s/p — y/qr=. Third product,
y/p + ^j = Fourth multiplier,
p — j' = Fourth product
We have this process for the reduction of \/p + %/q,
k/p + Vq = Given surd.
</i>' + C/g* — Vp'g — Vp^ + Vi> V = Multiplier,
— Vp*q—V^
p + q = Product
BXTBAGTION OF KOOTg. 269
^mmm ^^^^^ «^»i^
The </p + l/q can be rednced in a siiKiilar manner.
t^ + ?/y = Given expression,
i/p — Vfi^ = First multiplier,
i/p^ + </j'= -!/p^— 4^« Kret^prodact,
Vy— Vg'+ i/pq « Second multipKer^
i* -T 5? = Second product.
All expressions composed of two surds, whose common index is a
multiple of 2 and 8, may be reduced in a similar manner.
Tios, Ji/pj. !j/^^ Given expression,
v^F— ?/F*= Kist mulfcipKor,
VP — </7== First product,
t/p + t^7 = Second multiplier,
-i^p — v/g^ » Second product,
Vj>— </g+ </^ = Third multipKer,
JP — £ «■ Third product.
An expressions composed of two surds, whose common index is some
power of 3, may be reduced in the same manner as (/P + VF
Take as an example
Vp + K/q'= Given expression,
^I^ + </g^~ </^ « Mm multiplier,
Vy + V? = 1/F+ </£= First product,
i<P'+ Vg*— V^ a: Second multiplier,
p + gf = Second product.
Take as a second example, Vp"+ ?/^
Then, ?J/^ + %/Y — Given expression,
VF + X/¥— \/pq = First multipKer,
V^+ V? = Vp + Vq = First product, which can be
reduced as before.
The rednotion of 1/^ +VHf is more difficult than any of the pre-
ceding reductions.
260 FORMATION OW THS P0WSB8 AND
We will write the terms of tlie product that cancel each other in the
same vertical column.
y/P + V? = Given expression,
y?+ ^T— ^¥q— ^Pg^+ -y pV+ ^gy — ^W = MnltipKer .
P + ^"A+<? + W— -JW— ^^" + ^^+ ^W
— y7g — y;)?* + ^pV+VpV—:!/p^— ^Wf
p + q = Product.
All expressions composed of two monomial surds, may he rendered
rational when the surds have a common index. The amount of diffi-
culty attending the reduction depends altogether upon the index.
When the index is some power of 2, the reduction is very easy. But
when it is 7, 11, 13, 17, &c., the reduction is difficult.
When an expression is given to be reduced, we must first examine
the factors of the common index, and make our reduction correspond
to those factors.
Thus, let it be required to render rational !{/p'+ x/qT The factors
of 15 are 3 and 5 ; we must then reduce the expression by the first
multiplication, so that the common index of the result shall be 5.
Thus, v^p + y/q =s Given expression,
w/y« + !J/5»"— \/pq= Firet multiplier,
»y^+ '^5* =± </p + ^/q, which can be reduced as before.
To reduce the \/p+ \/qy we must use such a tnultiplier as willleave
a common index, 5, in the product.
Thus, VF + Vly_
Vp-^ql
\/p* — v^?' = ViP — ^qy which can be made rational as
before.
Let it be required to reduce ^/p + %/q.
The factors of the index are 3 (2)'. Therefore, we must first get
rid of the factor, 3, and reduce the expression to a common index,
(2)» or 8.
SXTRACTION Of BOOTS. 261
•j/p -J- "J/^ sss Given expression^
^^ + iy^—^pq B First multiplier,
Vi' + t^9 ^ ^^^ product,
^p — ^q = Second multiplier,
t/p — Vs^ ^ Second product,
Vi> + i/Sl ^ Third multiplier,
^p — ^q ^ Third product,
-s/p + -%/? = Fourth multiplier,
p — g z= Fourth product.
Corollary,
318. The principles developed in Case II. enahle us to find the ap-
proximate value of a fraction, whose denominator consists of a mono-
mial surd connected with known terms, or of two monomial surds.
Let it he required to find the approximate value of — = =z
_,. 2 2(>/5 — v/2) 2(223 — 1.41) 2(0-82)
'7r:r72 — 5^=^2 - 3 ^-3 —
1-64
— g— to within -01.
•J
EXAMPLES.
3
1. Required the approximate value of — = — to within -Ol.
^8 -I- -\/6
Am. *59.
3
2. Bequired the approximate value of = to within 'Ol.
8 -|- v^5
^'"- "59"'
3 _ 3(8 — ^5) _ 3(8 — 2-23) _ 3(5-77) _ 1731
*^^'8 + ^6"" ^-^ ~ ^9 -^69--"59-
3. Reanired the approximate valne of = to vithin '1.
4+1/2
Ant. 12'6
262 rOBHASIOM 01 THE rOWKBS AND
For QQ - _^ _ ^ 66 . ( y(4)« +_V (2)«— y (4)^2)
"'4+y'2 yF+ya (y4*-f4^xy(i7"+y(27-y(4»)-2)
66 ( (4)' + 16 - 5) ^
(4)' + 2 - "•
2
4. Eequiied the appioximate yalae of — = =. to within -1.
«/4 + t/2
^n«. -7.
For 2 ^ 2(i/(4)' + </(2)'— </4-2) ^ 2(2-5 + 1-6 — 2
4-2
5. Required the approximate value of =^ to within -01.
11— y/5
Am. 13-23.
6. Required the approximate value of — = =. to within -01.
Vll + >/6
Ans. 1-08.
7. Required the apmozimate value of - 1^ to within -1.
Ant. 14-4.
30
8. Required the approximate value of — = ==. to within -1.
^ ^25 4- V6
Ans. 6-4.
g
9. Required Ae approximate value of — ==^ r:. to within -001.
\/6— V5
^n<. 37-44.
2
10. Required the approximate value of — = to within -1.
^4—4^2
^lu. 6-1.
CASEin.
To make rational an expression containing three or more terms
of the square root.
819. Let it be required to make rational ^/p + ^/q + ^/n.
xxTR^OTioN 01* aoo». 26B
The ivooess is ua fbUows :
^P + v^? + v^» = Oive^ expression,
■^■^^ ^^w ^^^^ •
v/j> — v'y + ^n =5 First multiplier,
— \/pq — ^/qn — q
+ \/pn + y/qn + n
P — q + n + 2y/pn = First product,
p — qi- 7^—ly/^ ^ Second multiplier,
{p — q-Vnf — 4jpii » Second product
Let it be required to make rational y/p + ^/^ — ^'^.
The process is as follows :
>/p + Vq — \/n = Given expression,
y/V-^- s/q + s/n = First multiplier,
P + y/pq — -v/p»
+ Vi?? + q — y/qn
+ \^pn -f y/qn — n
P + 2' — n + 2 v^ = First product,
P + q — « — 2 v^jjjr =sr Second multiplier,
(P + fi' — »)■ — 4pgr = Second product.
Take as a third example y/p — y/q + y/vp + y/n.
^P — y/T + y/rn + y/'n
% /p + y/q ~ ^"^ 4. ^^
P — Vpq + y/pm + y/pn
** + >/pn — ^nq + Vnm
— q + >/pq + >/ w^ + y/^
- ^ — y/pm — y/mn+ y/qm
(j> + n — q — rn) -f 2y/p^ + 2y/^ = First product,
(p + n — q — m) — 2y/'^+ 2^~^ ^ Second multiplier,
^ + i^y/q^ — 4p» + 4^ = Second product reduced, and
in which P represents
the rational term.
264 FORMATION OF THB POWB&S AND
We may represent P — 4p» + 4qm by M", and tlien we irill have
M* + 4Pv/jm = Second product,
M' — 4P v^gm = Third multiplier,
M* — 16Pjm = Third product.
Corollary.
320. The principles of Case III. enable us to approximate to the
value of a fraction containing three or more monomial surds.
EXAMPLES.
1. Reduce — = rr= — to an equivalent fraction havine a ratioDal
denominator. Ans, 6 + \^6 — -s/SO.
For _ ^^_ -. _^ 10(v/5 - v/6 + 1)
1/5 + V6 + 1 (n/5 + n/6+ 1) (^/5 — V6 -fl)
10(v/6— ^/6 + 1) _ 10 VlCx^S"— v^6"+ 1) _ _
^7f 2(5) =5— v'SO + >/5.
The approximate value of the result can be found, if desired.
2
2, Reduce —= 1= zr=. to an equivalent fraction with a ra-
x/e + vs+x/M
tional denominator. Ans. (x/6 — y/S 4- \/T4"(6 — x/84)
=48 •
20
8. Reduce = = =ir to an equivalent fraction with a
1 + ^5+^6+ >/10
rational denominator.
Ans. (1 — v'5— n/6 + >/10) (n/10 + ,/30)
g
4. Reduce — = = = =. to an equivalent fraction with
^1 + v^2 + v/3 + %/4 ^ ^ _
a rational denominator. Ans. (3 — ^2 — -v/3) (4 + 2^6).
EXTBAOTIOK OV BOOTS. 265
2
5. Reduce — = =: = =zto an equivalent fraction with
\/3 + v'S + ^6 + \/8
a rational denominator.
^^ (v/8"— v^5"— s/6 -f .^8) (^24 + ^/S0)
— 6
2
6. Reduce — = -= = = to an equivalent fraction.
v/3 + v^5 + v^6 + v^lO_ __ _ _
Ans. -v/3 — -^5 — -v/6 + v'lO.
EXTRACTION OF THE SQUARE ROOT OF A MONOMIAL SURD
CONNECTED WITH A RATIONAL TERM, OR OF TWO MONO-
MIAL SURDS.
321. Before we proceed to extract the root^ it will be necessary to
demonstrate three principles upon which the extraction depends,
FirU Prin^ple,
The square root of a quantity cannot consist of the sum of two parts,
one of which is rational and the other irrational.
For, if possible, suppose >/a = re + y/'y. Then, by squaring^both
members, we get a *= «* + 2x>/y + y. From which, ^/y ==
s ' That is an irrational quantity equal to a rational one,
which is absurd. But the absurdity has not resulted from an error in
the analysis, and must, therefore, be in the condition.
Second Principle,
When the first member of any equation contains a monomial surd,
connected with rational terms, and the second member is made up u^
the same manner, the rational quantities in the first member are equal
to the rational quantities in the second, and the irrational quantitioB m
the two members are respectively equal also.
Let a + Vft s=: a? + >/y, then a = a, and y/h sa y/y.
For if a be not equal to x, let a = a; zt: m. __Substitute Jhis ^^ <*
in the equation, and there results x±«t+ ^6ssx-f V!/>^^^ .
+ y/h^ Vy^ which is impossible, (principle first.) Therefore, itis
absard to suppose that a is unequal to x. Hence, a =» ^>_^ —
equation becomes a + V^=s a + \/y^ or by cancelling a, >/ 6 = V !/•
23
t
1 /
266 FORMATION OF TKS P0WBR8 AND
Third Principle.
If v/a + v'^== ^ + >/yy ^^ ^^ v/« — s/^= « — v/y-
For, by squaring the first equation, we get a + s/h = as^ + 2x^}f
+ y(l).
From which a = cc* + y (2), and ^/h «= 2ajv^y (3).
Subtract (8) from (2), and we have a — -v/6"= x" — 2a;^y"+ y (4).
Extracting the square root of both members of (4), we haye
Y^ a — s^T= X — -/y .
This last equation has been correctly deduced from the equation
v^a + ^h =7 X + v^y; so that it is a true equation, prctyided that the
equation from which it is deduced is true.
322. The foregoing principles enable us to deduce a formula for the
extraction of the square root of an expression made up of two mono-
mial surds, or of one monomial surd connected with rational terms.
Let, k/ a + -v/^== » + v'y. (1).
^n>en, \/a — v^T= a; — v^yT (2).
Squaring (1) and (2), there results
a + v'Tss a^ + 2a5v^jM- y. (S).
a — y/h = a:" — ^x^/y + y. (4).
Adding (3) and (4), and cancelling (2) in the result, we get
a = a?'\-y. (6).
Multiplying (1) and (2), we get ^a* — b = x*— y. (6).
Adding (5) and (6), we get « + v^»' — ft. _ ^ (^^
Subtraoting (6) fiwm (5), we get ^ s= y. (8).
Extracting the 0quare roots of both members of (7) and (8), and
there resulte ^ a + ^/a« — fe _^ ^g^
IZTBAQTXOII Of HOOTS. 267
«d \/ "-^/-' = ^7- (10)-
By adding (9) and (10 ), we get the value of so + y/y^w of the
equivalent, \/ «+ v^*
By subtracting (10) from (9), we get the value of as — y/y^ or the
equivalent \ / a — y/Sl
Hence we have, by performing these operations, the two formulas,
sja V- >/ ' + >^- - i_ ^ ' - •y^ , (B).
By using the double sign =b, we may unite (A) and (B) in one
formula.
We will now show the use of the formulae.
EXAMPLES.
1. Let it be required to extract the squajre root <tf 9 + >/\h. Then,
by comparing 9 + >/45 with a + \/&, the quantity whose root is to
be extracted in (A), we see that a =& 9 and 6 s= 45.
H«ice, sJa + v^= y^9 + sfU^^J^ + ^^^^-^ ^
. /9 — yfW^h _ _
V 2 •= >/7}+ y|:
A result that ean be verified ; for, squaring it, we will have, 7} +
2. Required the square root of 9 + \/72. By eomparisoo, we get
a = 9 and & := 72.
268 FORMATION OF THE POWEBS AND
And from (A) we get k/^ + y/1^ = ^9 + V81 — 72 ^
This result can be verified ; for, squaring it, we get 6 + 2 V 18 + 8
= 9 + ^^727
3. Required the square root of 9 — ^56.
Then, a = 9 and 5 s 56. Formula (B) must be used.
From (B) we get \/9 — ^/'m = \/ ^ + ^^^
— 56
2
\/
Squaring this root, we have, 7 — 2^U + 2 = 9 — ^66,
4. Required the square root of 9 + 2\/ — 162.
Pas s the 2 u nder the radical , thus, 9 + 2 >/— 162 = 9 + >/— 648,
and >/a« — 5 = V81 + 648 = -/T^ = 27, and the applicatiou of
the formula will give K/d + 2>/^=n.^ = ^18+ V^==^.
This result, when squared, will give 18 + 2 s/ — 162 — 9 sb 9 +
2 V — 162. Hence, the result is true.
5. Required the sum of \/9 + 2y/'^^SZy and \/9—2^ — 162.
This result can be verified ; for, 2 >/ 18 squared = 72 ; and t he given
expression, when squared, produces 9 + 2>/ — 162 + 9 — 2 >/ — 162
+ 2 V81 + 648 = 18 + 2 >/ 72r= 18 + 54 = 72.
6. Required the square root of >/8 + ^/ — l*
^«.V^^+\/
V8 — 3
This result can be verified by squaring it. We will get "^ +
V8 — 3
+ V^ - ^^+ ^^^-
XXTEACTIOK OF BOOTS. 209
7. Required the square root of 1 + %/ — 15.
Verify the result by squaring it
8. Required the square root ot 1 + 2 ^ — 20.
Ans. V 6 + y/ — 4.
Verify the result by squaring it.
9. Required the square root of 1 + ^ — 288.
Ans, 8 + ^y—Sl
Verify the result by squaring it.
10. Required the sum of v/l+ %/— 288, and \/l— ^— 288.
An$, 6.
We see that the square root of an expression containing an imagi-
nary quantity, cannot have all its terms real. But the sum of the roots
of two expressions, involving imaginary quantities, may be real and even
rational. The formulas (A) and (B"^ are generally applied to expres-
BiouB for which >/a' — b is rational.
11. Reduce — = = to an equivalent fraction with a rational
V9 + t/4
denominator. Ans. 10(v^'3" — \/2)»
For J^ ^ ^ 10 (V(9y - t/(^y - V(9)'4 + V9(4)') ^
'4/9 + ^4 9—4
= 2 (^"27 — s/W— v^"l8+ v'T2)=2(3v'3"— 2v/2'— 3^2+2v'3)
^ 2 (py/W— 5v/2) = 10 (v'F— v^2).
Or, the reduction may be performed thus,
(/9 + V4 5/3 + ^2
(^3 + v/2) (>/3 _ v/2 3 — 2 ^^
4
12. Beduce — = = to an eqaiTalent fraction with a rational
denominator. Ans. 4 ( (2 + </S)(i/^—2y/Y'— v/10).
23*
270 roBHATioN or tbx povxbb and
Jf or, — T= =. ^ 4 c 3 ^
_V5 + V4 _ _ 5 — 4 _
4(^/6 V5 — -v/S — ^/lO + 2V5) =. 4((2 +^/55(V5)— 2^/2 —
13. Reduce --= =. to an equiyalent fraction with a rational
denominator. Ans, |(yT+ «/^— 1)2).
For _^ ^ 5 -^ (^(^^' + ^(*^^' - ^(^> ^^)
Vl + V2_ ( V4 + V2) (^(4/ + ^/c^y — V W (2) )
= I (2^2 + ^4 — 2) = I ( V4 + ((/2 _ 1)2).
14. Reduce — = = to an equiyaleni fraction with a tatioDal
^27 + V8
denominator. An$, |.
For _^ = 6 (V(27)' + V(8)' - V(27) (8) ) ^
' V27+</8 85
A(9 + *-6) = « = «.
The result is rational because the given expression was reaDj
rational; though under an irrational form.
16. Reduce =. to an equiTalent fraction with a lational de-
n + ^m
nominator. . m (n — %/m)
n" — m
IMAGINARY QUANTITIES.
828. Any expression whatover, made up of monomial surds and
rational terms, or monomial surds only, may be rendered rational by
repeated multiplications. The few examples given will show the man-
ner in which these multiplications must be made. No general rule can
be ^ven in regard to them. There is a particular class of monomial
surdS; which, when rendered rational, present some differences from
other surds in regard to their algebraic signs. These are imagioaiy
surds.
An ima^nary quantity has been defined to be the even root of a
negative quantity, because no quantity, taken as a fSactor an even sum-
ber of times, can give a negative result. Thus, y^ — a, 1/ — a,
^— a, ^ — a, are ima^aiy quantities.
IXTBAOTION or BOOTS. S71
If the indicated foot be of the 2m^^ degree^ then "V — ^ T'^J 1>o
written %/ct*^ — 1 = bi/ — 1. In which, h represents the 2f»*^
root of a, whether that be rational or irrational. And we see that all
imaginary quantities may be decomposed into two factors, one of which
is an indicated even root of minus unity, and the other of which is
real, and sometimes rational.
This decomposition mnst always be first effected previous to opera-
ting upon imaginary quantities.
To square ^ — a, we first write the expression ^a ^ — 1 ; these
two factors will both drop the radical when squared. Hence, (v/ — a)'
= (V'a ^/"=^7=- (a) (-1) = — a.
So, also, (v/— a)' = (^a ^— Vf = (ay/a) (— 1 ^Z— 1) « —
(^_ a)*=(v/« >/— l)*=(aO (+ 1) = + a\
(^— ay^(y/a s/— !/=(«' ^/a) (+v^— l)=aV— a.
(v/- a)«=(v'a y/- l)«-(a») (- 1) = - a\
The table shows that the even powen of imaginary monomiala are
always real, and that their signs are alternately plus and minus.
824. A table of products inD show the modifications of their alge-
braic signs.
(+•— a)(+v/-a) = -K-Za y/— 1) (^/a ^— 1) -. (^a)
(.•«)(v'-l)(^/-l)— a.
(—%/—«) (— >/— a)=(— v/a</— 1) (— v/av/— 1)=(— ^a)
(—>/«) (v/— 1) (^— 1) = — o.
(+•— g) (+ •— 6 ) - (^/oy— 1) (^/6 v'— 1) - (</«)
2T2 FORMATION OF THE P0WEE8 AND
(4.V-a) (— V— 6)«(^/a^/— 1) {—^b V— 1) = (y/a)
We see that like signs produce minus, and unlike signs produce
plus, when imaginary monomials are multiplied together.
825. We will now form a table of quotients.
+^—b "" -v/^v^^=^ "" +x/b
H- v^ — a v/a \/ — 1 ^a
—y/—b " —y/b v'^-l ~ s/b
— ^ — a ^ — y/a v^ — 1 — ^/a__ y/a
— ^ — a — x^a \/ — 1 — y/a
And we see that, in division, like signs produce plus, and unlike
signs produce minus. It would seem that, since the rule for the
signs in division is different from that in multiplication, the product
of the quotient by the divisor might not give the dividend. But any
of the preceding quotients can be readily verified.
x/—b +y/b _
+ y/a + ^a
Multiply the quotient -r by the divisor ^/ — b, Then^ -t
+V'^ __ _
X y/—b = - — jiy/b v^— 1 = +y/a s/—l = y/—a, the
+ V o
dividend.
826. Imaginary quantities can be operated upon just as real quan-
tities, provided that care be given to attribute the proper algebraic signs
to the results.
Let it be required to divide 4y/ — J" by 2b y/ — 1. The division
cannot be performed until 4y/ — 6' is transformed into the equivalent
expression, iby/ — 1, the quotient will then be 2.
Let it be re quired to m ultiply 4^/ — 6' b y 2by/^^. Then,
4v^— 6« X 2by/— 1 = 46 v'— 1 X 2,by/~ 1 =r — 8^.
BXTSAOTION 09 BOOTS. 278
Now let it be requir ed to divide — 86^ by 2hs/ — 1. Then,
2h^— 1 26^/— 1
EXAMPLES.
1. Multiply + %/ — a by — ^ — b. Ans, + ^/ah.
2. Divide ^ah by y/ — a. Ata, — v^ — 6.
y/ab _ y/as/h _ >/h __ — ^/^ y/^^^y/^^^
y/hy/— 1 = — y/^^^.
The minus sign is placed before the square root of h^ in the expres-
sion, — y/h y/ — 1 y/ — 1, iu Order to make the result positive.
Since ^/— 1 -vA^ = —1.
3. Divide y/ab by — y/ — b. Ans. + ^ — a.
4. Multiply aby/ — c* by >/ — a*6c. Ans. — a'6c v^ftc.
5. Divide — a^be^bc by v' — »*^- -^««- + <»^ >/ — !•
6. Divide — a^y/bc by ai^ — c". -4n». + a V — ^.
7. Multiply — V — ohc by — a5c V — cLbc- -^'m. — a*W.
8. Divide — a'6V by — >/ — a&c. iin«. — ahc ^/— aftc.
9. Divide — a'W by — a6c y/ — abc. Am, — y/ — ahc.
10. Add together a>/ — 6* and 5^/ — a*. -4n<. 2a5\/ — 1.
11. Kequired the square power of a + by/ — 1.
Ans. a« + 2dby/— 1 — V.
12. Required the square root of a' + 2a&^/ — 1 — V.
Ant, a + 6>/ — 1.
13. Bequired the third power of a + &^/ — 1.
Ans. a* — 3afi» + Ba'by/— 1 — 6V— 1.
8
274 FORMATION OV THB POWIBS AND
14. Required tte product of o 4- ^V — 1 *"id a — by/ — 1.
Ans. a* + V.
15. Required tlie quotient of a* + 5' by a + h^/ — 1.
Ans. a — hy/ — 1.
16. Required the quotient of a* + 6' by a — by/ — 1.
Ans, a + by/ — 1.
17. Required tbe cube root of a» + 3o«6V— 1 — 3a5'— yy^— 1.
Ans. a + by/ — 1.
18. Required the fourth power of a + 6 V— 1.
Ans. Q> + 4a»6 ^/.— 1 _ 6a«6« — 4a&» V— 1 + fc*.
19. Subtract 25/— 2 from 10^/— 2. -4»». 8>/— 2.
20. Add together 8-/— 2 and 2^— 2. ^n«. lO-y— 2.
21. Multiply V — a + y/ — b by %/ — a — >/ — b.
Ans. b'^-a.
22. Divide b — a by ^ — b + y/ — a.
Ans. — y/ — b 4- V— a.
28. Multiply J + 2^— a by J — 2>/— a. Ans. I + 4a.
24. Divide i + 4a by J + 2^— a. Ans. i — 2^— a.
25. Multiply 2 V— a by 3 v^^ a. ^»«. 6 %/— a %/— «.
26. Divide 6 V— a 4/— » by 2^/— a. ^n». 3 V— c
27. Divide 6>/— a (/— « by 3\/— «• ^»w. 2\/~ «-
28. Multiply (/— a + t/_ 6 by V— « — 4/— 2».
Aivs. 'J — a — ^/— ft.
29. Multiply 1 + by/— 1 + c>/_ 1 by 1 — 6>/— 1 + c>/— 1 .
-4w«. 1 + 6» — c« + 2c^/— 1.
30. Multiply 1 + by/—\ + cv/— 1 by 1 + ft^/— 1 — c/-- 1
Ans. l + c» — i» + 25v/— 1.
XXTBAGTION OF BOOTS. 275
31. Multiply 5v/— 1 + >/— 3 by 3 v'— 1 — %/— 27.
Am, —15 — 3 v'^^- 15v/8'-f-9.
32. Required the second power of 5^/ — 1 + ^ — 3.
uin«. — 25— lOv^T- 3.
33. Beqniied the square root of — 25 — 10-v/3~ — 3.
An9, 5\/ — 1 — 3.
34. Divide 6 4- ^ — 4 by 2 + ^ — 9, and make the denominator
of the quotient rational. . 18 — 14v/ — 1
Ant. ^-s .
For ^+-^"^ = (6 + ^ — 1) (2 — ^119) ^
2 + v'— 9 (2+ x/— 9) (2 — ^—9)
12 + 2^^^^^— 6^-^:^^^^^ ^"STy^^^ v^^^
4 + 9 ""
12 + 4>/— 1 — 18v/— 1 + 6 _ 18 — 14x/— 1
13 "*" 13
35. Divide 2 + V — 2 by 2 — >/ — 2, and make the denominator
of the quotient rational. . (2 + sf—ty 2 + 4^/1I^2
iliM. -^ 3 or .
36. Divide 1 + V — 1 by 1 — ^/ — 1, and make the denominator
of the quotient rational. Aim. v^< — 1.
37. Required the quotient of 6n/ — 4 divided by 2%/ — ^
Afu, 2.
An examination of the foregoing results will show some properties
of imaginary quantities.
1. Beal^ and even rational results^ may be obtained by the ordinary
algebraic operations upon imaginary quantities.
2. Two monomial imaginaries raised to an even power^ will always
give a rational result.
3. Two monomial imaginaries multiplied together, or divided, the one
by the other, will always give a real result
4. A monomial imaginary connected by the sign, plus or minus, with
one or more rational terms, and then rabed to any power, will give at
least one imaginary term in the result.
5. Any expression involving one or more monomial ima^nanes, may
be rendered real by one or more multiplications.
276 EQUATIONS OF THK SECOND DEQREE.
EQUATIONS OF THE SECOND DEGREE.
827. An equation, inTolviDg a single unknown quantity, is said to
be of the second degree, when thb highest exponent of that unknown
quantity in any one term is 2. Thus, a;^ -f x = m, and x* + o = i»,
are equations of the second degree with one unknown quantity.
An equation, involving two unknown quantities, is said to be of the
second degree when the highest sum of the exponents of the unknown
quantities in any term is equal to 2. Thus, ay = m, a;* + y == w, and
y* 4- a: =p are equations of the second degree with two unknown
quantities. For they may be written^ x^y^ == w, a^y -f y = «, and,
j^x^ -f a: = w, and, therefore, come within the definition.
We will begin with equations of the second degree, with a single
unknown quantity. These are divided into two classes, complete and
incomplete, A complete equation of the second degree^ with a single
unknown quantity, is one which contains the second and first powers
of the unknown quantity. Thus, a:" + a; = 0, c7x' + x = m, and
ax* -f- 6x = c, are complete equations of the second degree, with one
unknown quantity. There may or may not be other terms, besides
those involving the unknown quantity. The first and second powers
of the unknown quantity may or may not have coefficients dilTercnt
from unity.
An incomplete equation of the second degree is one containing the
second power only of the unknown quantity, and it may or may not
involve terms in which the unknown quantity does not enter. The
unknown quantity may or may not be affected with a coefficient diffe-
rent from unity. Thus, ax* = x*, x"-fm = », -j-4-c = — q are
incomplete equations of the second degree. Equations of the second
degree, with one unknown quantity, are frequently called quadratic
equations, because the unknown quantity is raised to the second
degree, or squared.
INCOMPLETE EQUATIONS.
328. We will first examine incomplete equations.
The general form of such equations is ox* j- <i = e. By
transposition and reduction; we get (ca — i)x" =s ce — of, or c* =
EQUATIONS OF THB SEOONB DEORBE. 277
— i ~ = gr, by substitnting for the known terms in the second
member a single term, equal to them in value. Hence, eveiy incom-
plete equation may be reduced to two terms, and the equation be
placed under the form of a^ =s q. Owing to this circumstance, incom-
plete equations are sometimes placed in the class of binomial equations j
or equations involving but two terms.
There is no difficulty in solving the equation 7^ = q. The square
root of the first member will give us x ; but if we extract the square
root of the first member, we must extract that of the second also, or
the equality will be destroyed. Extracting, then, the root of both
members, we have dbx = dz y/q. We have prefixed the double sign
to both members, because this equation, when squared, must give back
the original equation, x* = ^, from which it was derived, and either
-f- X or — Xj when squared, will give -f x^, and either + y/q, or
— y/q will, when squared, give q.
The value of the unknown quantity, or the root of the equation, as
it is generally called, has been defined to be that which, substituted
for the unknown quantity, will satisfy the equation, that is, make the
two members equal to each other. We see that an incomplete equa-
tion of the second degree has two values, numerically equal, but
afifected with difiierent signs. Either value or root will satisfy the
equation, for the result of the substitution of either *Jq, or — y/q for
X, in the given equation, will hQ q=zq. We have prefixed the double
sign to both members, but it is usual to prefix it only to the root. In
that case, however, we must understand that the sign of x is not neces-
sarily positive, it being affected with the positive sign only, when it
corresponds to the positive root. It becomes, then, i>ecessary to have
some notation to distinguish the values. This we will do by dashes.
The X that corresponds to the positive root we will write a/ ; and the
X that corresponds to the negative root we will write a/'. The first is
read x prime, and the second, x second. It will be shown that a com-
plete equation has two roots, generally unequal in value. We will
ako distinguish these roots or values by writing them x', and a/'. Some
equations of high degrees have three, four, &c., values. These will be
written a/, x", x'", x>^, &o., and read x prime, x second, x third, x
fourth, and so on.
To solve an incomplete equation, it must first be put under the form
of X* = gr. That is, all the known terms must be transferred to the
second member, and the coefficient of x* must be made plus unity, if
24
278 EQUATIONS OF THX 8BOOND DSOBEK.
not already so, by dividing both memberB by the coefficient of o^.
After tho equation has been put under the form of 7? = q^ extract
the square root of both members.
EXAMPLES.
1. Solve the equation, 2x' + 1=4.
Transposing, and dividing by the coefficient of x', we get 2^ = |.
Hence, x = ±>/|. Then, a/ =+ v/|, and a/' = — ^/ J. The solu-
tions may be left thus, or we may extract the indicated root approxi-
matively.
2. Solve the equation, 2x' 4-4 = 1.
Reducing, we get x=i:\/ — |. Then, a/=+^/ — J, and a''
These roots are imaginary. How are we to interpret them ? An
imaginary quantity indicates an impossible operation. Ought not the
equation which produces it involve an impossibility ? In this instance,
23(^, an essentially positive quantity, is added to 4, and their sum is
required to be less than 4. The condition of the problem is, then,
absurd, or impossible, and the result is impossible, as it ought to be
It will be shown more rigorously, hereafter, that an imaginary solution
always indicates absurdity in the conditions of the problem. The ima-
ginary values, though they fail to satisfy the conditions of the problem,
yet will satisfy the equation of the problem ; as they manifestly ought
to do, since they have been truly derived from it. Substituting either
+ ^/ — I, or — v^ — I, for Xy in the equation, 2a:*+4=sl, we get — f
+ 4 = 1, or 1 = 1.
329. We observe, then, this remarjcable analogy between imaginaiy
values in equations of the second degree, and negative values in equa-
tions gf the first degree. The values, in both cases^ will satisfy the
equation of the problem, but wOl not satisfy the required conditions.
There is, however, this difference : to convert a negative solution into
a positive one, numerically equal to it, we have only to impose a single
condition ) but, to convert an imaginary solution into a real solution,
precisely equal to it numerically, we have generally to impose two
conditions. Thus, the equation which gives the imaginary solutions,
expressed in words would read thus : required to find a number, twice
the square of which, augmented by 4, will give a sum equal to 1.
BQUATIONS OV TH£ SECOND DSaEXX. 279
The yalues are a/ =r + ^/ — |, and a^' = — ^ — |. Now, to get
Toal Taluee numerically equal, and not to make any change upon the
arithmetical yalues of the known quantities in the equation, it must
be changed into 2a? — 4 = — 1. This equation, expressed in words,
would read thus: required to find a number, twice the square of
which, diminished by 4, will give a difference equal to — 1. The
yalues of the new equation are a/ =: + >/ + |j and cc" = — >/+ 1-
These yalues are real, and differ from those in the first equation only
in the signs of the quantities under the radicals. To get these new
yalues, we made no change, arithmetically, upon the numbers in the
first equation, but we made two changes of sign, or, in other words, we
imposed two conditions upon the equation of the problem.
*
8. Solye the equation, r£^ + & =r a.
An9, a/=-fv^a — b^af'^z — Va — i*
Now, let a = 4, and 6 = 6. Then, 0/== + ^— 2, and a/'=: —
^/ — 2. To get the real yalues, a/ = + V2, and a/' =z= — V2, we
inust make h and a interchange yalues. h must be 4, and a, 6; that
18, two conditions must be imposed.
7?
4. Solye the equation, -^ — 1 + 7? — 4+ 5 = 3x* — 7.
Am. x'=: +2, a/' = —2.
5. Solye the equation, -g- — 1 + a? — % + 6 = 3x« — 22.
Ans. a^= +3, a/' = — 8.
6. Solye the equation, ^5 — 1 +af — 16 + 5 = 3a;' — 48.
Am. aj' = + 4, x" = — 4.
7. Solye the equation, |a;« _ ^ + 4x» — 20 + 100a:« — 500 «
995 — 199x«. Am. 7!^+ s/b, a/' = —^T
8. Solye the equation, J+ J _|._8+7a:«— 42 + 999a:'= 5994.
Am. o' = + ^\ a/' = — -v/6r
The preceding examples are simple, and the solutions can be readily
obtained. But there are many of a more complicated character, and
of course more difficult to solye. No general rules can be given to aid
the student; he must exercise his own ingenuity. It is well, howeyer,
280 SQUATIONS or THE BEOOND DSaBEE.
to make the clearing from fractions the first step in every redaction,
and then, if there is a single radical in the equation, it ought to be
placed in one member by itself.
EXAMPLES.
1. Solve the equation, -r 1- 8 = -j .
Clearing of fractions, we get ^x^ + 144 + 12 = re".
Placing the radical by itself, we have y/x^ -f 144 = re" — 12.
And squaring both members, there results sc" + 144 ^ x*— 24a:^ +
144. Hence, x* = 25x", or x* = 25. Then, x'= + 6 and x'' = — 5.
X 10
2. Solve the equation, = ^r^.
^ X -h v/x« + 44 22
Then, 22x = lOx + 10 v'x' + 44, or 12x = 10 ^ x* + 44.
Squaring both members, we get 144x" = lOOx" + 4400, or 44x* ^
4400. Hence, x' = 10 and x" = — 10.
8. Solve the equation, i + 5 = x*.
c c
These results can be verified by substituting either the value of of or
x" for X in the given equation. Then we will have Kj — -^ f- ftV
=. c(?^-6), or, by squaring both member, 1+^ + W
Hence the equation is satisfied.
• X (
4. Solve the equation,
X + ^a^ + a a
An*, oi = — _ , x" = — .
BQUATIONS OF THE SSOONB DEORBS. 281
Verify ihese resnlts. What do they become when a = 26 ? Why 7
What, when a = 0? Why? What, when 26 > a ? Why?
6. Solve the equation, v'jp' -f a:" -|- a; =5 mx.
^m' — 2m ^/m* — 2m
Verify these results. What do they become when m = 2 ? Why ?
What, when m = ? Why? What, when m = 1 ? Why? What,
whenp = 0? Why?
When there are two similar radicals, it is best to unite them in the
same member.
6. Solve the equation, = — ,
h + my/l — x' «
Ans. 0/= -f ^a'K-^2 + 2an6 (6-m) ^, ^ _
no — am '
^o« (m» — 6*) + 2anb {b — m)
w6-^ afn
For, clearing of fractions, we get nb + nhy/1 — a^ = ah -{■ am
^1 — x", or, (rib — am^ x^l — a;^= b (a — »), or, (nb — amf
(1 — ««)«ft»(a— ny.
Developing, we get nW — 2nbam + aW — a^ (n6 — amy == 6V
— ^2an6' 4- Vn\ Hence, a' (m' — 6") + 2an6 (6— m) = 3[^(nb — amy.
Hence, .' = + ^^ZME^+^^OEE:^, ^' = _
' (no — am) '
V g' (m' — 6') + 2an& (6 — m)
n6 — am
7. Solve the equation, n>/j>' -f x" + aa = x^p^ + x" + an.
.4n«. a/ = + v^a* — l>*i a^id a/' = — V ^* — JP*.
Verify these results. What do they become when a' = p' ? Why ?
What, when jp" > a« ? Why ?
330. There is another value which does not appear, it is x = n.
For, by the second principle, Article 321, the rational terms in the two
members must be equal. Equating them, we get ax = an, or a; = n.
This ought to be so, for the given equation can be put under the form
of (n — x") y/j^ -fa;* — a (n — a?) = 0, an equation which can
plainly be satisfied when x = n. The given equation, previous to the
24*
282 XQVATIONS OT THE 8E00ND DSORMS.
diTisioii by the oommon factor, n — Xj was really a cabio equation, and
contained three values. We are sometimes enabled to detect a yahie, as
in the above example, by equating the rational factors. When the root
of this equaticm will also satisfy the equation formed by equating the
radicals, it is, of course, a value in the given equation.
8. Solve the equation, pVx" — a* •{- a=. ny/7^ — a^-\-x.
((p — ny-f 1)
An9. 2B = a, or X = a -^f^- — 7 — r~.
1 — (p — n)*
331. There are some expressions in a fractional form, which must be
changed into equivalent fractions with rational denominators.
n a 1 xi. *• ♦» + «+ \/2fna; + cc"
9. Solve the equation, - = «.
m + X — -x/ 2mx + x"
2(2 +V^ 2(2— >7iO
For, by multiplying numerator and denominator of the fraction in
(m -f 05 ■\-^2mx H- rr*)*
the first member by the numerator, we get
«»«
Clear the equation of fractions, and extract the square root of both
members. Then, m + a: + y/lmx + «■ = dr fn>/n, or m (1 rb -%/«)
— aj = — x/1mx + cc*. Squaring both members, we get m" (1 =b 'Jnf
— 2fmao)\ ± >/ n) 4- a:* =r 2ma; + a", and, by transposition, m* (1 ± %/••/
= 2mx(2d=>/^. Hence, x = '^^^^ "(!J !-
-_ _ . _ , ^x* -f- a' + ^J 7^ — ^
10. Solve the equation, — ^= =z==r:=l.
Making the denominator rational as before, we have
^-^^^t^^l^^^^
Then, 2v'a;* — a* = 2a"— 2x», or >/a:* — a* = a:* — a«.
Squaring again, we get a:* — a* = aj* — 2a:«o" + a*, or 2a;«a* = 2o*,
or a:* = a' Hence, a/ = + a, and a/' =: — a.
It will be seen that these values satisfy the equation. In this ex-
ample, the second step was not to extract the square root of both mem-
bers, as in the kst example, because the double product of the radicals
SQT7ATXON8 Of THE SECOND 0BQRBS. 283
gave a simple result. When the radicals are of such a form that their
doable product will not give a simple result, it is best to make the
second step the extraction of the square root of both members.
11. Solve the equation, — ^ ===^ = 1.
Am. a/= + h,af'z=:z — h.
12. Solve the equation, - : — = _ - = n.
v/-c* + 8m«— -v/x* — m"
Ans, a/ = + w, x" =: — m.
Sometimes the first step is squaring both members.
13. Solve the equation, = = </ — .
Ans. a/ = H v/2an — m*, a:" = y/2an — m*.
n n
14. Solve the equation, ^a + x + y/a — x = ^2a.
Ans. a/ = + a, ai/' = — a.
16. Solve the equation, — = =s — .
ui?M. x'aai + 2^am — m\x^'^ — 2^ am — w".
What do these values become when m^^al What, when t» ^ a ?
../. « 1 .t .- 2x -f- 4 aj — 1
16. Solve the equation, — — y = . .
jIm. aj' « + >/5i aj" « — V 6l
... « , , . 2x + 4 X — 2
17. Solve the equation, ^ ^ == ^^^^ ^
Ans. x' =1 + 2, «" =« — 2.
What does the second member become when the first of these values
is substituted ? How do you explain the jesult.
4/NM1 t .• a; + a x + 6
18. Solve the equation, « .
^ X — h X — a
Ans. Both values infinite.
How are these results explained ?
284 EQUATIONS OF THS SECOND DSGBBS.
nx -|- a X ^ h
19. Solve the equation, =- = .
^ X — nx — a
^ ^ id^ — y^ „ fa^ — V
Wbat do these values become when n = 1 ? Why f What, when
2x + 1 a; + 2
20. Solve the equation,
X
— 2 ""20; — 1
Am, ^= + >/— 1, a/' = — >/— 1.
Why are these results imaginary ?
^, _ , , . .a hx — o
21. Solve the equation, aX'\ = — = a.
X XT — X
2 8a; 2
22. Solve the equation, 2x + — = -j| 2.
a^ ic ^^~ X
Afiz, x' = + 2, aj" = — 2.
3 27a; 3
23. Solve the equation, 3a; H = —= 3.
X XT — X
Aru. a/ = 4- 8, a;" =5 — 3.
PROPERTIES OF INCOMPLETE EQUATIONS.
332. 1st. Every incomplete equation of the second degree has two
values, and but two, and these values are equal with contrary signs.
For, the general form of the equation is x" = j or a:* — y = 0. The
first member may be regarded as the difference of two squares, and can,
therefore, be placed under the form of (x — -s/ j) (x -f- %/ j). Hence,
the equation, x' =: y, may be written (x — Vq) ^x + >/ j) = 0. Now
the product of two factors being equal to zero, the equation can be
satisfied by placing either factor equal to zero. Therefore, x — y/q = 0,
and X 4- ^q = 0. From which, we get a; = -f x/ j, and a; = — Vy,
or, distinguishing the values by dashes, a/ = 4- ^q, af'=z — ^q.
Since there are but two factors, the equation can be satisfied in but two
ways. Hence there are but two values, and we see that these are equal
with contrary signs.
By solving directly the equation, a;' = ^, we would obtain the same
results. But the process we have adopted is to be used hereafter for
SQUATIONS OF THB 8E00ND DBOBEE. 286
complete equations, and it is well to know that the properties of oom-
plete equations are also those of incomplete equations.
2d. Eyeiy incomplete equation of the second degree can be decom-
posed into two binomial factors of the first degree with respect to x ;
the first factor being the algebraic sum of x and the first value with its
sign changed, and the second factor being the arithmetical sum of x
and the second value with its sign changed.
The fectors have already been obtained, and are (x — Vq), and
(x + >/?) ; ai^d we see that these correspond to the enunciation of the
second property. The product of these factors is zero. Hence, when
we know the two values of an incomplete equation, we can always form
the incomplete equation itself which gave those values. We have only
to change the signs of the values, and connect them with x. We will
then have the binomial factors, and, placing th^ir product equal to zero,
we will have the equation required. Thus, form the equation that
gives for x the two values, + 2 and — 2. The two factors are (x — 2)
and (x + 2), and the equation is (x — 2) (x + 2) = 0, or x" — 4 = 0.
The result can be verified by solving the equation, a^ — 4 = 0. Or,
since in the equation, (x — 2) (x + 2) = 0, we have the product of two
&ctors equal to zero, the equation can be satisfied by placing either
factor equal to zero. Hence, x — 2 = and a; + ^ = 0. And these
equations, when solved, give the values, + 2 and — 2.
EXAMPLES.
1. Find the equation that gives for x the two values, + a', and — a'.
Atis. x' — a* = 0.
2. Find the equation whose values are + Vab, and — >/a&.
Arts, x" — ab = 0.
3. Find the equation whose values are + a — b, and + b — a.
Ans, x'+2ab — l^ — a^ = 0.
4. Form the equation whose values are + y/^ — a, and — V m — a.
Ans. a^ + a — w = 0.
5. Form the equation whose values are + >/m — a — v^m* — a",
and — V m — a-\- v^m* — a'.
Ans. of — (m — a) — (m' — «")+ 2v/m — a >/fn? — a' = 0.
Verify this result by solving the equation.
286 MQUATIONS or THB 8E0017D BBOESX.
6, Solve the equation whose yalues are a — x/a* — m*, and — a
+ v'a* — m*. Ans. a:* — 2a« + m" + 2a^/m" — a* = 0.
Verify this result by solying the equation.
7. Form the equation whose values are -f — j and
__ s/m« + of
8. Fonniheequationwhosevahie8are-^^,and- "^^^
Ans. (m* — fiF)c^r=.afr^.
9. Fonn the equation whose values are m' — n'; and — m* •!• n*.
^fw. a:» = «n* — 2mW + «*.
What do the values become when m-=.nl What^ when m = ?
BINOMIAL EQUATIONS.
333. Any binomial equation of the n*^ degree oan be solved as the
preceding equations. We have only to extract the n*^ root of boA
members.
Let X* = 27; then, x = V^= 3. Let x» = q; then, x ^ss. yq,
K n be an even number, the radical must have the double sign, and
two values, at least, of the equation will be known. If n be an odd
number, only one value will be known. It will be demonstrated here-
after, that every equation of the n^^ degree has n values. The method
of finding the other values will then be explained.
Let a?* = a"; then, x a* db^/a* ss db\/ v'c? = dz^ifc a.
Then , x' = + v^T^ x" « — v'T^ ^" = •^y/'-^^a, a/'" s —
v/ — a.
In thiB example the four values of x have been determined, but it
is not often the case that they all can be found.
Let »• = a*; then, a; = ± ^1? = ±t^ ^lof = ± l/dtz a, and
four of the six values are known.
EQUATIONS Ot THE BXOONP PSGBSS. 287
Let »*■ = a*; then, x = db!;/? = dcv/v^ = ±^'«; and only
two of the twelve yalaes are known.
XXAHPLES.
1. Solve the equation, ^x^ — 6az + 3a* a 2 — 2a.
An$. d rsjfv^zjt^a, a/' -s — -^/^ + a, a/" = + v'— (« + a),
a/''' = — ^/— (2 + a).
2. Solve the equation; x^ = 256. An&, of ^b + 4, a/' 9 — 4.
3. Solve the equation, ^a* + a;* + \/a' — x" s mv^2.
^fw. a/ = + (/a* — (wi* — a')', »" = — (/o* — (m»— a«)*.
4. Solve the equation, x"*^ ss 32.
^iw. a/ = + ^2^ x" = — -^2^
6. Solve the equation, v^i>' + ^' +y P' — ^' =pv^"Z:
Am. a/ = -{-pt/m (2 — m), x" = — pt/m (2 — w).
6. Solve the equation, x^ = 1.
An9. af= + 1, a^' = — 1, x^'dr +^^=T, x"" = — v^^^T
Yerify the preceding resulto bj substituting them in the given
equations.
GENERAL PROBLEMS IN BINOMIAL EQUATIONS.
334. 1. The sum of two numbers is a, and the ratio of their squares
m : what are the numbers ?
An,. a^ = ±^i^ ^=Z:±>^
1 + v^m 1 — ^m
For, let X = one number; then, a — x ==the other, and, by the
a"
conditions, -^ r-^ = m. Now, extract the square root of both mem-
' (a — x)* ^
bers, then, = db v^m. Hence, -, == + ^^m, and -j.
a — X a — a; a— ar
= — ^n. From the first equation, x' = -f av^m — x' \/»w, or
288 EQUATIONS OF THE SECOND DEOBEE.
a/(l + v^w) = a \/fn, or a/ == =. From the second equa-
1 H- v/m
tion, we get a/' = — a ^la + ^'^'niy or a" = ^— . Now, if
1 — \/f»
m = 1, tbe second value is infinite. How is this result to be inter-
preted ? By going back to tbe equation of tbe problem, we see that
it becomes, under tbis bypotbesis, p ^, = 1 ; an equation wbicb
can only be true wben a = 0, or wben xz=z—. Tbe first supposition
contradicts tbe enunciation, tbe second is tbe value found for a/. The
second value can only exist when there is a contradiction to the state-
ment : it ought, then, to appear under tbe symbol of absurdity. We
may explain tbe result otherwise; thus : wben m = 1, the equation in
a/' becomes x" — a" = — a, or j-, — = -f -^j, or = + -^.
This equation is plainly absurd for any finite value of x".
Tbe value, a;" = — -rr-, satisfies the equation of the problem, as it
0* (H
manifestly ought to do. We have = 1, or -r;; rs- = 1,
^ ^ / a\ (Oa — a)* '
or, j]==^??^^l^%rOV = 0»(Oa — a)«, orO=0.
2. The sum of two numbers is a, and tbe ratio of their fourth powers
is m. What are the numbers ?
Ans a/- + ^^^ x" ^ ^^
1 + V*^ 1 — v»
3. Two numbers are to each other as m to n, m being greater than n,
and the ratio of their squares is equal to a' divided by the square of the
greater. What are the numbers ?
an ^ , . a»*
Am. First, ± — , Second, ± — r
4. Two numbers are to each others as m to n, and the sum of their
fourth powers is equal to a^. What are the numbers f
Aw, First, ± - , Second,
yn* + w* i/n'' + m*
EQUATIONS OF THl SECOND DSOBXB. 289
5. Two nmnberB are to each oiher. as m to n^ and the difference of
their fourth powers is a*. What are the numbers ?
AiKs. First, ± J Second,
What do these values become when m z^ nf Why? What when
f»»sO? Why?
6. The cube root of twice the square of a number is 2. What is
the number ?
Atim. Either + 2, or — 2.
7. The cube root of a times the square of a number is h. What is
the number ?
Am.af=^+sJ^l^,x"=.-sJt.
What do these values become when a s ? What when 5=0?
8. The square of a number multiplied by the first power of the same
number is equal to 64. What is the number ? Ans, 4.
9. A man put out a certain sum of money at 6 per cent, interest.
The product of the interest upon the money for 6 months by the interest
for 4 months was 600 dollars. What was the sum at interest ?
Ans. $1000.
. 10. A man put out a certain sum of money at 6 per cent, interest.
The product of the interests upon it for 3, 6 and 9 months was $20}.
What was the sum ? Ans, $100.
11. The successive quotients, of a quantity divided first by a, and
then by h, will, when multiplied together, give a product equal to mn.
What is the quantity ?
Ans. x' = + \/ahmn, a/' = — y/ahmn.
COMPLETE EQUATIONS OF THE SECOND DEGREE.
fjljj^ COR
335. The most general form of such equations is, 7- + »=r;
clearing of fractions, we get, — am:^ + ^x^ — ^^^ » hmr. Dividing
TfCQc I)n J)T
by the coefficient of o?. — am, we have a? 1 = ; and
am a a
hex O Cth I V't
by transposition, of = ^ ^. Now, the second member
am a
25 T
290 EQUATIONS OF THE SECOND DEGREE.
may be plainly represented by a single letter, — q, and the coefficient of
the second may be represented by — p. Hence, the given equation as-
sumes the form of x* — px ^ — q; in which the highest term of the
unknown quantity has a coefficient plus unity. Had the coefficient,
— r- of ac" in the original equation been affected with the positive sign, it
would have reduced to the form of x' +px = q. Had the coefficient
of a^ been positive, and that of x negative, the equation would have
assumed the form of x^—~px as q. Had the last conditions been ful-
filTed, and the prevailing sign in the second member at the same time
negative, the equation would have assumed the form of x' — px= — q.
And, since every change that may be made upon the signs of the
coefficients of x' and x, and upon the signs of the known terms, will
eventually lead us to one of the preceding forms, we conclude that
eveiy complete equation of the second degree may be placed under one
of the following forms :
x* + px = ly. First form,
a^ — pxssi ^q^ Second form,
x' + px = — q, Third form,
x' — px = — q. Fourth form.
When a complete equation is given to be solved, it must first be placed
under one of these forms. To do this, clear it of fractions, if it con>
tain any, and then make the coefficient of the first term plus unity , if
not already so, by dividing by this coefficient. The rcsnlting equation
will then be under one of the required forms. The form that the
resulting equation assumes will, of course, depend upon the prevailing
signs in the given equation.
x* Ix
Reduce the equation, — ^- + 3= — |,to one of the four forms.
Clearing effractions, we have x* — 2x+12s= — 6, or x* — 2x:^ — 18.
Hence, the equation has assumed the fourth form.
oj* \x
Reduce the equation, -j ^+ 3 = — ftoan equivalent equation,
which will appear under one of the four forms.
Multiplying both members by 20, the least common multiple of tiie
denominators, we will get 5x' — lOx + 60 = — 24. Now, divide by
•f 5, to make the coefficient of x' plus unity. Then, x* — 2x + 12
24
_-.^or-4|.
EQUATIONS OF THE SEOONP DEGREE. 291
We see in this particular case, that we might haye obtained the same re-
sult by multiplying the original equation by 4. This would have made the
coefficient of the first term unity, which is the main point to be attended
to ; then, by transposition, the equation would have become :^ — 2x =
— 16|. And we see that the equivalent equation is of the fourth form.
It frequently happens that it is impossible to make all the terms entire,
and the coefficient of the first term at the same time plus unity. But,
since this coefficient must always be plus unity, we derive the following
rule for reducing any equation to one of the proposed form.
RULE.
MuUyply both members of the equation by the reciprocal of the ca-
efficient of the first term^ and then transpose all the known terms to
the second member.
The multiplier in the equation, fee' — 2x = 4is, + |. The multi-
plier in the equation, — | cc" — 2aj = 4 is, — |.
The reason of the rule is obvious, and needs no explanation.
EXAMPLES.
1 . Beduce the equation, fa^ — ^a5-|-6 = 4, to one of the four forms.
Ans. 05* — ^= — 3.
2. Beduce the equation, — fa^ — i^ + 6=-f- 4, to one of the
four forms. . • , a?
Ans. a:« + 5 = + 3.
3. Beduce the equation, -^a^ + i^xz=lyto one of the four forms.
Ans. a^ + bx ^ 26.
4. Beduce the equation, ~ — - =: 1, to one of the four forms.
Ans, 35* — 6x = 26.
6. Beduce the equation, — fic* — ^cc -f-6 = -|-4, to one of the four
Ans. a:" -{- 5 = 3.
336. K the first member of the equation put under one of the four
forms is a perfect square, the solution can be as readily effected as in
the case of an incomplete equation. For, we will only have to extract
292 EQUATIONS OF TH£ SECOND DEGREE.
the square root of both members^ and then transfer the known term or
terms in the first member to the second member, and then the solntion
will be complete. Suppose that the equation is x* + 2ax + a' =&.
Extract the root of both members, then x + a=idb s/b. Hence,
a/ a — a + \/b, and of' z= -^a — V t. We see ihat the equation
has two yalues; and that these are not numerically equal, as in the caae
of incomplete equations.
Now if, by any artifice, we can make the first member a complete
square, it is plain that there will be no difficulty in the solution of a
complete equation of the second degree. Let us then assume the equa-
tion, a^ +px = qy and examine what modification the equation must
undergo, in order that its first member may be made a perfect square.
The square of a binomial is composed of the square of the first term,
plus the double product of the first by the second, plus the square of
the second term. The square of a binomial is, therefore, a trinomia].
j^ + px must then be made a trinomial by the addition of some quan-
tity before the first member trill become a perfect square. What is the
quantity to be added ? Take the expression, (a -|- a)* s= rs* + 2ax + a*.
We SCO that the third term of the trinomial is the square of the second
term of the binomial in the first member ; and we see that 2aXf the
second term of the trinomial, when divided by 2a;, twice the first term
of the binomial, gives a quotient, a, which is the second term of the
binomial. This quotient, squared, is the third term of the trinomial.
Now, suppose we only knew the first two terms of the trinomial, z^ and
2aXf and wished to ascertain what was the binomial, which, when
squared, would give these for the first two terms of its square. We
would know that the first term of the binomial must be x ; and, since 2(ur,
the second term of the trinomial is twice this first term by the second
term, it is plain that the second term can be found by dividing by 2x, twice
the first term. Having found a, the second term of the binomial, we
have only to square it, and the third term of the trinomial will be
known.
Apply these principles to the expression, x" -f px, regarded as the
first two terms of a trinomial that is a complete square. It is plain
that X is the first term of the binomial, the first two terms of whose
square are x* + px. The second term of the binomial can be found
by dividing px by 2x, twice the first term : the quotient, -^, is the
second term of the binomial, and its square, ~, is the required third
EQUATIONS OF THE SECOND DEGREE. 293
term of the trimonial. If then "^ be added to x* +px, the first
member will be a perfect square. But^ if we add ^ to the first mem-
ber^ we must also add it to the second, else the equality of the two
members will be destroyed. Hence, the equation, cc!^ + j)x =s q, can
be changed into the equivalent equation, a^ •}• px -^ C.=:^-f^,
in which the first member is a perfect square.
Now, extract the square root of both members,
a: + |- « =h ^/f- + q- Hence, a/ =s — |. + y/^ + q, and
'"— f-\/l^
Either of these values when substituted for x will satisfy the equation.
The first, when substituted, will give ( — 9"+ V/T +?)''*'
Beducing the first member we have, q ^^ q- Hence, the first value
satisfies the equation. The second, when substituted, gives
PSJ^ + 2+1- + ? — f —P\/j + ff « ?» o'f ? = ?•
Hence, the second value also satisfies the equation.
We have taken an equation under the first form, but it is obvious
that the principles deduced are applicable to equations under either of
the three other forms. Hence, for the solution of a complete equation
of the second degree, we have this general
RULE.
Reduce the equation to one of the four forms, by multiplying both
members by the reciprocal of the coefficient of the highett power of the
unknoum quantity, and by transposing all the known terms to the second
25* •
294 EQUATIONS OF THE SECOND DBOREE.
member. Next, add the square of half the coefficient of the finl power
of the unlcnoion quantity to both members j and then extract the square
root of both members.
Finally, separate the unknown quantity from the known quantities
by leaving it alone in the first member,
A common error with beginDers is^ to complete the square without
observing whether the coefficient of the first term is plus unity. But
it will be seen that the rule requires the first step to be the making of
this coefficient plus unity, if not already so.
EXAMPLES.
1. Solve the equation, — x* + IQx = 28.
Ans. a/ = + 2, x" = + 14.
2. Solve the equation, x* — 4x »= — 4.
Ans. x' = + 2, x" = + 2.
3. Solve the equation, x* + 4x « — 4.
Ans, x' = — 2, x" =s — 2.
It is generally best to reduce the terms under the radical to a common
denominator.
Take the equation, — ^ ax" — ex ^ — m. This, when solved, ^ves
C #971 Cr
X ss — --±4/ 1- -_ The two terms under the radical can be
2a V a 4a*
reduced to a common denominator by multiplying the numerator and
denominator of the first by 4a. If m had been divisible by a, then a
would not appear in the denominator of the first term, and the multi-
plier then would have been 4a'. The single term, into which all the
known quantities in the second member have been collected after the
coefficient of x' has been made plus unity, is called the absolute term.
7Tb
In the preceding equation, — is the absolute term. Then, to reduce the
a
terms under the radical to the same denominator, we multiply numerator
and denominator of the absolute term by either^ times the coefficient
of the second power of x, or 4 times the square of this coefficient.
This rule is only applicable when the reduced term in the second
member was entire, previous to making the coefficient of x' plus unity.
BQVATIONS OF THE SECOND DSGBSS. 295
4. Solve Uie equation^ 2x* — 3a; «= 5.
Ans, a;' = + 2^, a/' = — 1.
For, solving, we get, x = | ± ^/f + T*ff = I =t v/ f g + A = | :
J« + 2i, or — 1.
5. Solve the equation, 2a^ — 3x = 2.
Aru. ocf = + 2, a;" = — i-
For, solving, we get, a; = | =h x/T+'y^ = I =*= ^Z j^ = {=*=
|=:+2, or— J.
o. Solve the equation, — , H =—5-1 .
m c m c
Ans, a/ s= 4- a, a;" ssr — a .
c
337. These few examples are given to show how complete equations
can be solved. But the solutions can be better understood when some
of the properties of these equations are known.
First Property.
Every complete equation of the second degree has two values, and
bat two, for the unknown quantity.
For, resume the equation, x* -|- px = 5'. Complete the square of
the first member, and we have a:' + pa; + ~ = g^ -f ^. Now, the
first member being a square, the second member may be represented by
p«
a square also. Hence, x* + ^jx -f- t" =* ^'; <^' (^ + !>)' — m* = 0.
Now, the first member being the difference of two squares, may be re-
solved by the principle of the sum and difference into two factors.
Then the equation, (x + j?)" — m' = 0, will be changed into the equiva-
lent equation, (x H- p + m) (x + j? — m) = 0. And since, when the
product of two factors is equal to zero, the equation can be satisfied by
placing either factor equal to zero, we have, x-f|?-|-«i=:0, and
a:+j> — m = 0. From which we get x = — p — m, and x = — jp + «i ;
or, distinguishing the values by dashes, a/ = — p — wi, and x" = —
JO + m. Now, if we replaced m by its value, we would have the solu-
296 EQUATIONS or THE SECOND DBOBBB'.
lions previously obtained. We see diat there are two iralnes nmne-
rically unequal, and, since the equation, (x + p -{- m) (x + p — «i) = 0,
contains but two factors, it can be satisfied in two ways, and only in two
ways. Hence, there are two, and only two, values for the unknown
quantity.
Second Property,
338. The first member of every equation of the second d^ree can be
decomposed into two factors of the first degree with respect to x, the
.first factor being the algebraic sum of x and the first value of x with
its sign changed ; and the second factor being the algebraic sum of x
and the second value with its sign changed. The second member of
the equation after this decomposition will be lero.
The factors have ' already been obtained, and are (2: + p + ^) ^^^
(x -{• p — m). By comparing these factors with xf and a/', we see
that + p + w is equal to — a/, and that + p — wi is equal to — a^,
and these factors then can be changed into the equivalent, (x — xT)
and (x ^~ a/') ; and the equation, (x -{■ p •\- m) (x •{■ p — m^ = 0, can
be changed into (x — a/) [x — x") = 0.
339. This important property enables us to find the equation whose
values are known. We have only to change the sign of the first value,
and prefix -I* ^ ^ ^^; ^^^ ^^ ^^^ factor will be known ; and then to
change the sign of the second value, and prefix + a: to it, and the
second factor will be known. The product of these factors placed
c(|ual to zero is the equation required.
Find the equation, whose values are + 2 and — 3. The first factor
must be (x — 2) ; the second factor must be a + 3. Hence, (x — 2)
(x+3)=0, is the equation required. Expanding the first member, we
have X* + CD — 6 = 0, or x' -|- x b= 6. Completing the square, and
solving, we get x =: — i zh i/GTI =■ — i =h v/'^= — i dr | =
+ I, or + 2, or — | = — 3. Hence, the process is verified. But
the verification might have been made more readily, thus : the equation
(x — 2) (a; + 3) = 0, can be satis^ed by placing cither factor equal to
zero. Hence, x — 2 = 0, and aj + 3 = 0. These equations give the
preceding values, ■\- 2 and — 3.
A few examples will make the beginner more familiar with the second
property.
EQUATIONS OF THB SECOND PBQ&XS. 297
EXAMPLES.
1. Fonn the equation whose values are both — 2.
Ans. (x -h 2) (x + 2) = 0, or a:*+ 4a; + 4 = 0.
2. Form the equation whose values ore a + ^h, and a — y/b.
Ans. (x — a — ^b) (x — a+ v^i) = 0,orac* — 2ax=zb — a*.
3. Form the equation whose values are — 2m-{- Vn + 4w", and
• 2m — \/n + 4m^ Ans, a^ + 4kmx = n.
4. Form the equation whose values are a — n+ \/m — 2an -f n',
and a — n — x/m — 2an + n*.
Ans. X* — 2 (a — ji) x = m — a'.
Verify these results by solving the equations.
340. The second property of complete equations sometimes enables us
to solve an equation very readily. For^ whenever the factors of an equa-
tion are apparent, we need not solve the equation itself, but only those
factors separately placed equal to zero. Thus, take the equation
a^ + hxz=: ax. By transposition we get, ix^+ bx — ax = 0. We
see that x is a common fiictor to the first member, and the equation
may be written x (x-^ b — a) = 0. And since, when the product of
two factors is equal to zero, the equation can be satisfied by placing
either equal to zero, we have cc = 0, and a; + b — a = 0. Hence, the
two values are 0, and a — b. The general eguationj a? -{-bx — ax
= 0, shows, furthermore, that when the unJcnovm quantity enters into all
the terms of the first member of an equation, whose second member is
zero, one value of the unknovm quantity must always be zero.
Take, as a farther illustration of the use of the second property, the
equation, aa^ — ba? + ax — bx = bx — ax. This equation can be
written (a — b^a^ +2 (a — h)x = 0,ot x {(a — b)x + 2 (a — b) )
= 0, or a; (a — b) (x + 2) = 0, or a; (a — 6) = 0, and a; + 2 = 0.
Hence, a/ = 0, and a/' = — 2.
Again, a:* — aaj + ^u — a=»0, may be written x (x — a^ + x — a
= 0, or (x — a) (x 4- 1) = 0. Hence, a:' = a, and a/' = — 1.
We have seen that the first two properties of complete equations are
also properties of incomplete equations. The remaining properties are
also, as we shall see, common to both classes of equations.
298 EQUATIONS OF THE SEOOND DEOBES.
Third Property,
341. The algebraic sum of the valaes of eveiy complete equation of
the second degree is equal to the coefficient of the first power of the
unknown quantity, with its sign changed.
For the equation, x" + ^x = j, when solved, gave the two values,
Adding these equations, member by member, we get a/ + a/' = — />,
as enunciated.
The third property of complete equations is also a property of incom-
plete equations. For, an incomplete equation of the form, ocf ^ q, may
be written, x^ +0x sssq. The two values of this equation, solved
either as a complete or as an incomplete equation, are, a/ = + ^q^ and
x" = — \/q. The sum of these values, a/ -f- x", is plainly zero,
which is the coefficient of the first power of x. An incomplete equa-
tion then may be regarded as a complete equation, whose two values
are numerically equal, but affected with contrary signs.
Fourth Property,
342. The product of the values in every complete equation of the
form, a^ -\- px = qy is equal to the second member or absolute tenn
with its sign changed.
For these values are ci/ = — -^ + k/ x + ?•
and a:" ^-^^+q.
and their product, a/x" = + ^j 1^ + ^ 1 s= — q.
This property belongs also to incomplete equations, for the two values
of the equation, x* = g', are x' ^ -f- ^q and x" = — ^q, and their
product, x'x" = — q.
EQUATIONS OF THE SECOND DEOBBE. 299
We will 866; hereafter, that the third and fourth properties belong to
equations of every degree, and that the first and second; with some
modifications; are also properties of all equations.
EXAMPLES.
1. What is the product, and what the sum; of the values in the
equation, a;* + 4x = — 4. Ans. a/as" = -f- 4, a/ + a/' = — 4.
2. What ihe product, and sum, in the equation, a^ + x sal,
Ans. afa/' = — 1, a/ + a^' = — 1.
3. What the product; and sum, in the equation, a^ — px = — q.
Ans. a/x" = + gr, a/ + a/' = + p.
Fifth Property.
343. The value of x, in every complete equation of the second
degree, is half the coefficient of the first power of Xy plus or minus the
square root of the square of half this coefficient increased by the
absolute term.
For the equation; a;" + /^x = j, when solved, gives x = — ir"^
^/ ^ + q, which agrees with the statement.
It is obvious that this is also a property of incomplete equations,
since, for such equations, p =, 0.
We have shown the foregoing properties by operating upon an equa-
tion of the first form, but, since the demonstrations have in no way been
dependent upon the signs of p and q, it is obvious that equations under
the three other forms enjoy the same properties.
The fifth property enables us to solve an equation directly, without
completing the square in the first member; but it is well to require
beginners to complete the square, until familiar with the principles.
300 EQUATIONS OF THl 8E00ND DEOEES.
OSNEBAL EXAMPLES.
1. Solve the equation, aa^ -{• cx^s ma.
or.a/- 2- , OS .
2. Solve the equation, 7x* — 14a; = + 280.
Am. a/ = 1 + v^4i; x" = 1 — -^41.
3. Solve the equation, naj' — mx = nm.
J-. ^ , »» + V'wi' + 4n*m „ my/ir} + 4n^
^^•^==+ 2;^ '^= 2;;^ — •
4. Solve the equation, oa:* — 6a; «= ma;.
uln.. x' = 0, x" = !1±^.
a
5. Solve the equation, 4a;' — ^ 5a; « 7a;.
Am. a/ «= 0, a'' B= 8.
6. Solve the equation, ni? — mx = 5a:.
Am. a;' = 0, x'' = 5JtJ?.
n
How might the answers to 5 and 6 be deduced directly from the
answer to 4.
7. Solve the equation, — of' — px = -|- j.
8. Solve the equation, a;" + |x — 1 = 0.
uln«. x' -= 4- ^, x" ss. — 3.
9. Solve the equation, 4x' + lOx — 6 = 0.
Am. a/^ +^, a;" = —3.
10. Solve the equation, x* — ^x — j = 0.
Am. a/ = + 3, x" = — i-
EQUATIONS OF TH« SECOND DEORM. 301
11. Solve the equation, ic" -f » + /^ = 0.
Am. «' = — ^, a:" = _|.
12. Solve the equation^ x* — a; + -«-=: 0.
^n«. a/ = + I, a// =, + ^
13. Solve the equation, ICx" — 16a; + 8 = 0.
^n«. a/ =: + I, a/' = + |.
14. Solve the equation, o:^ — ax + ah=z + bx.
Arts, a/ = 4- a, a?" = + i.
16. Solve the equation, x^ + ax -^ ah = —hx.
Arts. x^:=z~a, a/' = —6.
16. Solve the equation, wia:« + — — ^ ^ '^^^
^n*. a/«^, a/' = _A.
^ a
17. Solve the equation, wx"— ^ -« ^ _ _ »««^
« b '
-4n«. a/ = + -1, aj" 5s _ ii
18. Solve the equation, aj«— ~— ???-. _^ iw:
6 « - ^— y
^««. a/ = 0, a/'= -?. + ^_c — ^
^ n g '
19. Solve the equation, a^ + ^+^^^xA-^
Ans. a;' = 0, a/' = c+^_^_.^
J b n'
20. Solve the equation, a:« + Twa; + ca: + ^U a>+ m'x = 0.
^71*. a/ = 0, a:" = — m — c — v^— m«.
21. Solve the equation, a^ — mx~cx — ^lix^m*x = 0.
^w«. a/ = 0, a;".= m + c + -•« + m*.
These equations show that, when the sign of the coefficient of x has
been changed, and the other terms left unaltered, the values will be
numerically the same, but affected with contrary signs. They show
26
302 EQUATIONS OF THE SECOND DEGREE.
also that, when all the tenns after o^ are negative, the second member
being zero, the values will be both positive ; and when the terms are
all positive, the second member being zero, the values will be both
negative.
22. Solve the equation, trx* '\- 'px — — 5.
^*'*- "^ = 2ik ' 2^^ •
What will these values become when ^=i\mq^? What, when
jp* ^ \mq f
23. Solve the equation, ^a? + 24a; = — 36.
Alls, ir' =r — 3, x" = — 3.
24. Solve the equation, 4x' — 24rr = -^ 36.
Ans. cc' = + 8, a^' = + 3.
25. Solve the equation, 4x' — 24x = —40.
Am. a/ = + 3 + \A^, a;"= + 3 — ^A=^.
The following important consequences flow from the last four ex-
amples.
Whenever the square of the coefficient of x is not greater than four
times the product of the coefficient of x' into the absolute term, the
values will be imaginary, if the sign of the absolute term is n^tiTe.
And, whenever the square of the coefficient of a; is equal to this na-
tive product, the two values of x will become identical.
344. Some equations may be treated cither as complete or in-
complete.
1. Solve the equation, ^ r. = w.
Am. X' = .^^, a/' = ^^.
1 + >/n 1 — v^ii
For, expanding the denominator and clearing of fractions, we have,
x' = n (a* — 2ax + x*), or (1 — w) x" + 2anx = na^^ or x* +
2an na?'
1 — n 1 ^ n'
Hence
^ an / ng' aV
— an-^ y/nc? (1 — n) -f aV — an + a,yn a^n (1 — y/n)
1 — n 1 — n ' 1 — n
Now, multiply numerator and denominator by 1 + y/n^ and we have,
EQUATIONS OF THE SECOND DEGBEE. 303
a^«(l-n)_^ g^^ In the «une way, a/' can be shown
(1— n)(l + v/n) 1+^n
equal to =. These yalues for a/ and x" are identical with
^ 1—y/n
those before obtained, when the equation, r ^ = », was treated as
^ (a — x)a '
an incomplete equation.
2. Solve the equation, ^ — -— — - = m", as a complete equation.
{a + x)
. , am „ — am
Ans, X = -^ , X ^
x«
1 + m' 1 + w'
3. Solve the equation, ■=- = g, as a complete equation.
(a: — slff ,
x/j — 1 ^/(? + l
345. In the foregoing examples, the unknown quantity has been
freed from radicals. Whenever it is connected with radicals, it must
be freed from them by the preceding rules^ and then the process will
be the same as in the examples already given.
1. Solve the equation, i/ma:"-f- nx = >/j>
n + \/n' + 4mp , n — v'n" + 4mp
2m 2m
2. Solve the equation, >/ mV + nx = ^^y as an equation of the
second degree. . , p „ — v
° Am, X = — ==- — , X = =— .
m + 71 m — — vi
3. Solve the equation, y/ma^ -|- wx = />j as an equation of the
second degree. ^^ ^ ^ _P , ^' J.
\/m + n y/m — ^
4. Solve the equation, -:, — —^ A =- = 3 J.
^ r+^^ x/l + x
-4w«. a/ =s + 8, a/'= — f
5. Solve the equation, 1 = m.
<* + ^ x/a + a;
^_ m'— 2(o+ l)+m^m»— 4 ^, _ m'— 2(a+ 1)— m^/
.A.7t9» X — • ' ' Q > — Q
m*
What will these values become when m = 2 ? What, when m <^ 2 ?
304
SQUATIONS OF THE SECOND DEGREE.
>/3 -f X 3 -f a;
6. Solve the equation, -s—r 1 ==
" + ^ v^3 H- X
Arts, a/
= — 2,a^' = — ±
7. Solve the equation, -^r— ; 1 z=:==
' l+x v^l + x
= 1.
-4n«. a/ =
—3 + %/— 3
2
tt
ft
_8_^_3
8. Solve the equation, =z = 1.
^/ 2cx -f- 4ac
What do these values become when c* = 2ac ?
9. Solve the equation, — = 1.
-a — ^/2ac — c*.
What, when c ]> 2a /
^8x + 32
jln«. a/ = — 2, a/' = — 2.
10. Solve the equation, — = 1.
y/lQx + 40
^7w. a/ = — 2 + ^— 6, a/' = — 2 — v"—
11. Solve the equation, ^a^ + bx= ^ax + y/hx.
Yerify all the preceding results.
Arts, x* = 0, x" = a + 2^ ah.
DISCUSSION OF COMPLETE EQUATIONS OP THE SECOND
DEGREE.
346. The four forms of these equations are
7?+pxz=zq,
Q? — pX = y,
/ 2c' + i>a: =" — qj
> a:*— px = ^2f,
and the corresponding values.
X
*f
First form.
EQUATIONS or THK SEOOND- OSaBEX
305
Second form.
Third form.
X' = :P +
2
v/-
rp-
S'^
^■=f-\/^-'.
Fourth form.
Now, we observe that the first and second forms differ only in the
sign of Xy and that of in the first form is the same as a^' m the second,
taken with a contrary sign ; x" in the first form is the same as a/ in
the second, taken with a contrary giign. A similar remark may be made
in regard to the values of the third and fourth forms. We conclude,
then, that to change the signs of the values, without altering them nu-
merically, we have only to change the sign of the coefficient of the first
power of X, Thus, x* + as = 2, when solved, gives a/ = 1, x" = — 2 ;
and x* — X = 2, gives x' = 2, x" = — 1.
In like manner, x* — 5x = 4, when solved, gives the two values,
a/ = + 4, x" = + 2 ; and x" + 6x = — 4, when solved, gives
x' = — l,x" = — 4.
If we proceed to extract the square root of ~-+ ^y io the two values
P
of the first form, we will get ^, phis a series of other terms. Henoe,
P P
the two values will become x = — 9 + ^ "'" ®*^' terms ; and as" =
— :^ _ ^ — other terms.
Hence, the first value is positive and the second negative, and the
second is numerically greater than the first, because the radical is added
to the quantity without the radical in the second value, and subtracted
firom it in the first value.
26* u
806 SQUATIOl^S OF THS SECOND DBOREB.
By extracting the square root of ^ + ^ approximatlTely in the
values of the second form, we would see that the first value was positiYe
and the second negative, and that the second was numerically greater
than the first. We have anticipated this, from what has been said
before, as to the interchange of position and the change of sign between
the values of the first and second form.
In the third form both values are negative, because — ~ is 2>
mi
n/
^ — q. The second value is numerically greater than the first
In the fourth form both values are positive, and the first is numeri-
cally greater than the second. These results might have been antici-
pated from our knowledge of those in the third form.
347. The discussion of the signs of the values in the four forma
may be made otherwise, thus : we know, from the fourth property of
equations of the second degree, that the product of the values in the
first form must be equal to . — <^. Hence, the values must have con-
trary signs. And we know, from the third property, that the sum must
be equal to — p. Hence, the negative value is the greater. For,
when we add^ negative and positive quantity together, and their sum
is negative, we know that the negative quantity is greater than the
positive.
For the second form, the product of the values is equal to — j.
Hence, the values have different signs. Their product is equal to -f- /?
Hence, the positive value is greater than the negative.
For the third form, the product of the values is equal to -f 9- Hence,
the values are both positive, or both negative. But their sum is equal
to — ]p. Hence, they are both negative. We cannot decide from this,
however, which is the greater, the first or second value.
For the fourth form, the product of the values is equal to + 9-
Hence, the values must have like signs, and must be both positive, or
both negative. But their sum is positive, being equal to -f p ; henoe,
they are both positive. We cannot decide from this, however, which ia
the greater, the first or the second value.
EQUATIONS OF THE SEOOND BEaBES. dOT
IRRATIONAL, IMAGINARY, AND EQUAL VALUES.
348. The values, in all the forms, will be irrational whenever the
root of the radical cannot be extracted exactly. In that case, the ap-
proximate value of the radical can be determined to within a vulgar or
decimal fraction, and then the approximate values of the unknown
quantity will be known.
There can be no imaginary values in the first two forms, because
the quantities under the radical are affected with positive signs. When-
ever^ then, the absolute term of an equation is positive, we know, cer-
tainly; that there are no imaginary values in the equation.
The third and fourth fctrms contain imaginary values, only when q is
t>' .
^ ^. For, in that case only, is the prevailing sign under the radical
negative. Whenever, then, an equation is put under the third or fourth
form (the unknown quantity being freed from radicals, and the coeffi-
cient of v^ being made plus unity), we can readily tell whether there are
ima^ary values. We have only to square half the coefficient of the
first power of x, and compare the result with the absolute term.
The equations, a:* + 10a; = — 30, and a* — 10« = — 30, both
contain ima^nary values, because (5)' <^ 30. Imaginary values, being
always similar in form, are said to enter the equation in pairs.
Since the two values of each of the four forms differ only by the
radical, it is obvious that the disappearance of the radical will cause
these values to become identical. But, the radical cannot be made to
vanish in the first and second forms, because the quantities under the
sign are positive. It will vanish, however, in' the third and fourth
forms, whenever ^=:q. The values of the third form wiH then both
become — ^, and of the fourth form + ^. The equality of the
values, in the third and fourth forms, differs from the equality of the
values of incomplete equations. In the latter, it is equality only in a
numerical sense ; in the third and fourth forms, absolute equality.
If we substitute ^ for q in the third and fourth forms, the equations
will become a:' -f ^a: = — ^, and a?* — px = — -j, or, by transposi-
tion, a^ + JMP H- ^ = 0, and ^ — px + ^ = 0.
308
EQUATIONS OF THE SECOND DEGREE.
The first member of one equation is the square of {x + =^), and of
the other (x — ^). Hence^ when there are equal values in the third
and fourth forms, the first members of those equations will be perfect
squares, provided the second members are zero.
The following are illustrations : x* + 6x = — 9, and x' — 6x =
— 9 ; x" + 8x = — 16, and x* — 8x = — 16.
SUPPOSITIONS MADE UPON THE CONSTANTS.
349. Known terms are frequently called constants, and unknown
terms, variables. In the equation, x* + px == g, jp and q are the con-
stants, and X the variable.
Let us make the constant ^ = o, in the four forms. By going back
to the solved equations, we ^ee that the values will become
ct' = ^ + -P = 0,
'^ - 2 2 - -P'
Finit fonn.
» - 2 ^- 2 ~ P'
Second form.
^ = -f + f = «'
V' _ P P - o
^ - 2 2 P'
Third form.
af'-A-P —P -0
a? - + 2 2 "'
Foorth fonn.
We see that the first and third forms have the same values, and that
the second and fourth forms have the same values. This ought to be
so, for the hypothesis, ^ = 0, makes the first and third equations iden*
tical, and also makes the second and fourth equations identical.
EQUATIONS OF THE SECOND DEGEES. 300
We have gotteu the foregoing series of values by making g = in
the solved equations. Ought we not to get the same results by opera-
ting upon the given equations themselves ? When ^ = 0, the first
and third forms become a^ + px = 0, or x (x + p) = ; an equation
which can be satisfied by placing either factor equal to zero. Hence,
X = 0, and a; + ^ == ; or, a/ = 0, and a/' = — p; the same as
before obtained. The second and fourth forms become^ when ^ = 0,
aJ — px = 0, or X (x — p) = 0. Hence, x" =: 0, and x' = + p, bs
before.
When ^ =r 0, we have this series of values :
j/ = +\/q, a/' = — x^q, in the first and third forms.
a/ = + x/ — q, x" = — v^ — q, in the second and fourth forms.
These results can be obtained directly from the equations ; for the
first and second become x' = q, and the third and fourth, od^ =s — q.
These two equations will plainly give the four preceding values.
When j3 = 0, and q = 0, the system of values reduces to a/ = -|- 0,
x" =s -|- 0, in each of the forms. This ought to be so ; for, in that
case, the four equations reduce to the same, x' = 0.
The solutions of the four equations give formulas which can be ap-
plied to particular examples. Thus, the values of the first form are
i^ = — f- + \/|- + ?» and af' = — ^ — \y^ + q. Let it be
required, now, to apply these results as formulas to find the values of
the equation, x' + x = 2. Then, p = 1, and q = 2. Hence, a/ =
— i + V^TT2"==— }+^=l. Andx" = — }— v/ir+2"= —
J — 1= — 2.
We have used as formulas the values a/ and a/', belonging to the
first form, because the equation, a;' -f x = 2, is of that form. And, of
course, we must always apply as formulas the values found in the form
to which the given equation belongs.
EXPLANATION OF IMAGINARY VALUES.
350. We have seen that the values of the third and fourth forms be-
come imaginary when q was made greater than ^. It remains to ex-
plain the cause of the imaginary results, and to ascertain what they mean.
In the third form, the product of the values is equal to + ^, and
their sum equal to — p. The values, then^ are both negative ; and
810 EQUATIONS or THE BEGOND DEGREE.
we have previously seen that they were unequal when ^ was unequal
to q. If they were equal^ each must be — ^. Let x represent the
ezecss of the greater over — =^. Then the value which is nnmericslly
the greater will be represented by — ( 9 + ^)> ^^^ ^^ smaller nu-
merically will be represented by — (^ — x\. Their product will
be "2^ x". Now, it is evident that this product will be the greatest
possible, when x = 0. This condition makes the two values equal, and
makes their greatest product -^^ . But q represents their product, and
it, therefore, can never be greater than -^. Hence, when we make
"j- ^ ^, we impose an impossible condition. An tmaginaiy ichuion^
therif indicates an impossible condition.
Ought not the equation to point out an absurdity when an impossi-
ble condition is imposed ? When the square is completed in the third
form, the equation is x* H-px -| x *° d ?> <>^ ^^ + 9 ) =t — 9-
Now, when ^ ^ 4"' ^^^ second member is essentially negative, whilst
the first, being a square, is essentially positive. We then have a na-
tive quantity equal to a positive quantity, which is absurd. A carejvl
inspection f then, of ike equation which produces imaginary values wHl
show an absurdity.
We have examined only the third form, but the preceding conse-
quences may be readily deduced by an examination of the fourth form-
The following problem will illustrate more fully the subject of ima-
ginary values.
Required to divide the number 10 into two parts ; such, that their
product will be equal to 30.
Let X be one of the parts, then 10 — x will be the other, and, from
the conditions we get x (10 — x) = 30, or — x' + lOx = 30. Mul-
tiplying both members by minus unity, we get sc^ — lOx eb — 30, and,
oomfdeting the square, x' — lOx -f- 25 ss — 6. Hence, x' = 5 -f
v/ — 6, and x" «b 5 — ^ — 6
GSNSBAL PROBLEMS. 811
The values are imaginary, as they ought to he; for the greatest pro-
duct that we can form of two numbers, whose sum is 10, is 25.
Hence, we have imposed an impossible condition, and the imaginary
values point out this impossibility. The equation, of^ — lOx -f- 25 =
— 5, may be written {x — 5)* ^ — 5 ; that is, a. positive quantity
equal to a negative one, which is absurd. The imaginary values will,
however, satisfy the given equation, as they ought to do, since they
have been derived from it.
EXPLANATION OP NEGATIVE SOLUTIONS.
351. Required to find a number whose square, augmented by three
times the number, and also by 4, shall be equal to 2.
The equation of the problem is x" + 3a; + 4 ss 2.
This equation, plainly, cannot be satisfied in an arithmetical sense ;
for 4 is already greater than 2, and must be still greater when aug-
mented by x' and 3x. Solving the equation, we get a/ = — 1, and
a/' = — 2.
These negative values satisfy the equation, but do not fulfil the con-
ditions of the enunciation. For, x being negative, the equation will
become x* — 3x -f- 4 == 2. The true enunciation of the problem, then,
is : Required to find a number, whose square, diminished by three times
the number, and that remainder increased by 4 will give a result equal
to 2. The equation of this problem, when solved, will give the two
positive values, 1 and 2.
Negative values then, in equations of the second as well as of the
first degree, satisfy the equation of the problem, but do not fulfil the
enunciated conditions ; and, since negative quantities indicate a change
of direction or character, those negative values point out the change
that must take place in the enunciation, in order that its conditions
may be complied with.
GENERAL PROBLEMS INVOLYING COMPLETE AND INCOM-
PLETE EQUATIONS OP THE SECOND DEGREE.
352. 1. Required to divide the number n into two parts, the product
of which shall be equal to m.
Am. x' = J n + s/W — »w, x" = in — \/in* — m.
What will these values become when m = \v? f What, when
w»>Jn«f Why?
812 GENERAL PROBLEMS INYOLVINO
2. Required to divide the number 10 into two such parts that their
product shall be equal to 25. Ans. a/ = 6, of' =z&.
Is the disappearance of the radical always connected with maximum
values ?
3. Required to divide the number 10 into two such parts that their
product shall be equal to 26.
Ans. x' = 5 -f >/-— 1, x" = 5 — V—l'
Why are these values imaginary ? Verify them by substitution in
the equation of the problem.
4. The sum of two numbers is a, and the sum of their squares is 6,
what are the numbers ?
An,. ^^« + vf^ a^' = "-^y^'.
When do these values become imaginaiy? when equal ? What is
the least value that the sum of the squares can have ?
5. The sum of two numbers is 10, and the sum of their squares is
52, what are the numbers ? Ans. of z=:Q, oc^' = 4,
6. The sum of two numbers is 10, and the sum of their squares is
50, what are the numbers ? Ans. oc' =5, re" = 5.
7. The sum of two numbers is 10, and the sum of their squares is
40, what are the numbers ?
Ans. x' = 5 + V — 5, a:" = 5 — >/ — 5.
How may the values, in the last three examples, be deduced directly
from the general values in example 4 ?
8. A Yankee pedlar bought a certain number of clocks, which he
sold again for m dollars. His gain per cent, on his investment was
expressed by the number of dollars in it. How much did he pay for
the clocks ?
Ans. a/ = — 50 + i/(25 + m)100, x" = — 50— >/(25 + m)100.
9. Same problem as the last, except that the pedlar sold his clocks
for 375 dollars, instead of m dollars.
Ans. a/ = 150, a/' =s — 250.
What is the meaning of the negative value ?
EQUATIONS OF THE SECOND DEGREE. 813
10. A number of partners in business owe a debt of m dollars^ but^
in consequence of the failure of one of tbeir number, each of the solvent
partners has to pay n dollars more than his proper proportion of the
debt. How many partners were there.
■ + \/v^' „ '-\/t+i
2 ' 2
11. Same problem as last, except that the debt was 728 dollars, and
that the increased portion of the solvent partners was 60| dollars.
Ans. a' = 4, a/' = — 8.
The negative value is readily explained. An examination of the
equation of the problem, after — x has taken the place of + x, will
show that there have been two changes of condition, and the correspond-
ing enunciation will be : '^ Several partners in trade owed a debt of 728
dollars, and, by the accession of another partner to share the debt with
them, their individual liability was diminished by 60} ; required the
number of partners.'' These changes in the enunciation give + 3 for
the number of partners, and the result can be verified. Bor, the share
of each in the debt is, 242} dollars before a new partner was added to
the firm, and but 182 dollars afterwards. And, in general, for problems
involving equations of the second degree, there must be two changes
in the enunciation to convert a negative into a positive solution.
12. The difference between the cube and the square of a number is
equal to twice the number. Eequired the number.
Ans. x'= + 2, a^' = — 1.
13. Required to divide a quantity, a, into two such parts, that the
greater part shall be a mean proportional between the whole quantity
and the smaller part. What are the two parts ?
Ans. Greater part, either » , or 5 ; and lesser
part, either |a — ^a ^/S, or |a 4- ^a v^6.
Both values will satisfy the enunciation, in one sense; for, a(|a —
4aVS) = ( -" + "^7 , and «(|« + ^a^-5) = (=^=^'.
Seoaose, by performing the indicated operations, we have for the first
27
314 QENEBAIi PB0BLBM8 INVOLVING
value, |a« — JaVl>"= X "" o" v^"^+ "T" = I*' "" 2" V^j and, for
2
"4
a' 5a*
tlie second value, |a* + ^a'v^= -^ + |a"v'T+ -p = |a" + Icf^T.
The second value, however, does not satisfy the conditions of the
problem, understood in a literal sense, for the greater part, — -^ —
-fTs/^j is less than the corresponding smaller part, |a + -^ \/?^ and
this corresponding part is greater than the whole quantity, a. The
explanation of the negative solution is simple, when we return to the
equation of the problem, 7^ 7ata{a — x). When x is negative, this
equation becomes, x" %= a(a -f- ^); ^^^ the corresponding enoncislioD
ought to be required to find a number, which shall be a mean between
the whole quantity, a, and the sum of a and this number.
The given quantity, a, might have been represented by a straight linei
B C A
^ j-y and the problem then would have
been to find a part, BO, which should be a mean between the whole, BA,
and the part, AC, left after BC was taken from it. Now, when the ex-
pression for BC became negative in the solution of the problem, it in-
dicated that BC must be laid off on the left of the point B, becsiiae
this distance was laid off on the right of B, when its expiesaioii was
C B A
positive. GUie diagram becomes ^ j-,
and AC is greater than AB. This agrees with the second value of AC,
3 a
•«■ ^ + o" v^^ which is greater than a or AB. The negative solu-
tion here indicates then a change of direction, and the equation being
of the second degree, there are two corresponding changes of condition.
The first is expressed by the unknown distance being sought upon the
prolongation of AB, instead of upon AB itself; the second is expressed
by the unknown distance being a mean between AB, and AB phu BC,
instead of between AB and AB minuz BC.
The explanation of a negative solution need never be difficult; we
have only to change -f x into — 2; in the equation of the problem, and
then examine and see what the resulting equation means.
Some of the following problems will involve one of the four methods
of elimination. Whenever the equations between which the elimiiia-
EQUATIONS OF THE SECOND DEGREE. 815
tion is to be effected is of a higher degree than the first, the method of
elimination by the greatest common divisor ought to be employed.
14. Find two numbers whose sum and product are both equal to a.
a —
-y/a^"
-4a
s/a^-
2
-4a +
7
a
y 2 '^ 2
When will the two numbers be equal ? When imaginary ?
15. Find two numbers whose sum and product shall be equal to 4.
Afii, 2 and 2.
16. Find two numbers whose sum and product shall be equal to 2.
Am, a/ = 1 + ^A=T, x" = 1 — -w/^=T, ^ = \— y/^^, f =
1 + ^/— 1.
17. Find two numbers whose sum and product shall be both equal
Do the values in problems 14, 16 and 17, indicate that there are two
distinct sets of numbers, or that the second set is the same as the first,
differing only in the position of the numbers.
•
18. Two capitalists, A and B, invested different sums in trade. A
invested half as much as B, and kept his money in trade 6 months,
and gained one-twentieth of his investment. B kept double the amount
that A had, in trade for 9 months, and gained 9100 more than A.
Supposing that their respective gains were proportional to their re-
spective capitals and the periods of investment, what were their
capitals ? Am, A's $1000, B's 82000.
Why was the second value of x rejected ? What was the gain per
cent of A and B ?
19. The difference between two numbers is a, and the difference be-
tween their cubes is h ; what are the numbers ?
^ - i^h — a*
^n.. a^=|-+^i^,a:" = |— ^^
12a
a , Mb — a" J, a Mh — a>
^ 2- + V-12^'^==^-2-V-12^-
When will oi* and y" be real, but negative?
816 GENERAL PR0BLSM8 INVOLVING
What do the negative values satisfy ? When will the two values of
X be equal ? How will the two values of y be in that casef When
will the solutions be indeterminate? How many conditions must be
imposed ?
20. The difference of two numbers is 2, and the difference of their
cubes 152 ; what are the numbers 7
Ans. First number, + 6, or — 4 ; second number, + 4, or — 6.
How are the negative values explained 7
21. The difference of two numbers is 4, and the difference of their
cubes 16 ; what are the numbers ?
Ans, First number, + 2 ; second number, — 2.
22. The difference of two numbers is 4, and the difference of their
cubes 15 ; what are the numbers ?
^ = -^-\/-^-
23. The difference of two numbers is zero, and the difference of
their cubes zero ; what are the numbers ?
Ans. a/ and a", both = -^, and y and y", also both a=s ~.
24. The year in which the translation began of what is called King
James' Bible, is expressed by four digits. The product of the first,
second and fourth, is 42 ; the fourth is one greater than the second,
and the sum and difference of the first and third are both equal to one.
Required the year. Ans. 1607.
25. The year in which Decatur published his ofBcial letter from New
London, stating that the traitors of New England burned blue lights
on both points of the harbour to give notice to the British of his at-
tempt to go to sea, is expressed by four digits. The sum of the fint
and fourth is equal to half the second ; the first and third are equal to
each other ; the sum of the first and second is equal to three times the
fourth, and the product of the first and second is equal to 8. Re*
quired the year. Ans. 1813.
26. The year in which the Governors of Massachusetts and Connect
ticut sent treasonable messages to their respective Legislatures, is ex-
pressed by four digits. The square root of the sum of the first and
EQUATIONS or THE SECOND DEOBBE. 317
seoond is equal to 3 ; the square root of the product of the second and
fourth is equal to 4 ; the first is equal to the third, and is one-half of
the fourth. Required the year. Ans. 1812.
27. A gentleman puts out a certain capital at an interest of 5 per
cent. ; the product of the interest for 12 months hy the interest for one
month, is just one fourth of the principal. What is the capital ?
Ans. a/ « 0, re" r= $1200.
28. Some of the New England States were fally, and some partiaUy,
represented in the Hartford Convention, which, in the year 1814, gave
aid and comfort to the British during the progress of the war. K 4 be
added to the number of States fully and partially represented, and the
square root of the sum be taken, the result will be the number of States
fully represented ; but if 11 be added to the sum of the States fully and
partially represented, and the square root of the sum be taken, the
result will be equal to the square root of 8 times the. number of States
partially represented. Required the number of States fully and
partially represented.
Ans, Three fully represented ; two partially represented.
29. The sum of two numbers is a, the sum of their squares b, and
the sum of their cubes c. What are the numbers ?
Have we two distinct sets of values, or but one set, with an inter*
change of position ? When will the values of x and y be equal ? When
imaginary ? When negative ?
30. The sum of two numbers is 7, the sum of their squares 25, and
the sum of their cubes 91. What are the numbers ?
Ans. a/ = 4, x" = 3; y = 3, y = 4.
31. The sum of two numbers is 6, the sum of their squares is 18,
and the sum of their cubes 54. What are the numbers ?
Ans. Both 3.
82. The sum of two numbers is 5, the sum of their squares 20, and
the sum of their cubes 50. What are the numbers ?
y=j + JV— 15.
27*
318 OBNEBAL PROBLEMS INVOLVING
88. Two tntyelleirs Btarted at the same time from two cities, C and
W; and travelled toward each other. They found, on meeting, that the
traveller from C had travelled 150 miles more than the other travdler,
and that by continuing at the same rate^ he could reach W in five days;
whereaS; it would take the traveller from W twenty days from the time
of meeting to reach C. Eequired the distance between W and C, the
rate of travel per day of the two travellers, and the time that had
elapsed before their meeting.
Ans. Distance between W and C, 450 miliss; the rates, 80 and 15
miles per day ; the time elapsed before meeting, 10 days.
84. The sum of three numbers is 15, the sum of their squares 93,
and the third is half the sum of the first and second. What are the
numbers ? Ans. 8, 2 and 5.
85. A man bought a tract of land for $11 per acre, and sold it again
at a less price, his loss per cent, on the sale being expressed by the price
per acre which he received. What did he sell the land for f
Am, 10 dollars per acre.
36. In the year 1637, all the Pequod Indians that survived the
slaughter on the Mystic River were either banished from Connecticut,
or sold into slavery. The square root of twice the number of survivors
is equal to j^^th that number. What was the number f Am. 200.
37. A Southern Planter bought m acres of cultivated, and as many
of uncultivated, land. He got h more acres of uncultivated' than of
cultivated land per dollar, and the whole cost of the cultivated exceeded
that of the uncultivated by c dollars. How much did he pay per acre
for each kind of land ?
Am. a^=~b ^ ^ ./ b(im + bc) , x" =
2 V /.
— b
X , /b(4m + bc) for cultivated land; y =
- 4- J . / b (4m + bc) y y" = i_^. / r(4w+ bc )^
2 V /. 2 V c
c
for uncultivated land.
88. A Southern Planter purchased 100 acres of cultivated^ and 100
acres of uncultivated land; the former costing him $500 more than the
EQUATIONS OF THE SECOND DEGBSE. 319
latter; the smaller cost of the latter resulted ^m his getting for eyery
ddlar y^th of an aere more of the nncnltiTated land than of the cultivated.
What was the cost per acre of the cultivated and uncultivated land.
Ana. Former, $10 per acre; latter, $5 per acre.
Yerification. 100 acres at $10 per acre will cost $1000, and 100 acres
at $5 will cost $500. A dollar will buy ^^^th of an acre of the culti-
vated, and Jth of an acre of the uncultivated land ; and the difference
between ^th and y^i^th is juth. So, a dollar will buy J^th of an acre
more of the uncultivated than of the cultivated land.
39. In the year 1853, a number of persons in New England and
New York, were sent to lunatic asylums in consequence of the Spiritual
Rapping delusion. If 14 be added to the number of those who became
insane, and the square root of the sum be taken, the root will be less
than the number by 42. Required the number of victims.
Arts, 50.
Why IS the second value of x rejected ?
40. Two farmers, A and B, invest each a certain amount in the
Central Railroad of North Carolina. After a time, A sells his stock
for $150, and gains as much per cent, on his outlay as B invested.
B also sells his stock and gets $12} more than he gave for it, but his
gain per cent, on his outlay is only half as great as that of A. Required
tlie amount invested by each. Aug, A, $100 ; B> $50.
Why is the negative solution rejected f
Verification. 50 per cent, on $100 is $50. And since A sold his
stock for $50 more than he gave for it, he gained 50 per cent, on the
$100 of outlay. B gained 25 per cent on his outlay^ and 25 per cent,
upon $50, is $12}.
41. Two travellers set out at the skme time, the one from A, and the
A B C
other from C, and travel towards -, -,
each other at uniform rates. After meeting at B, the traveller from A
is a days in reaching C, and the traveller from C, c days in reaching A.
How long was it after the time of starting until they met at B, and
Low long was each traveller in performing the distance A C.
Time of meeting, dc ^ac. Traveller from A, adz ^/ac; traveller
firom C, c db \/«c.
820 GENERAL PROBLEMS INVOLVING
What is the meaning of the negative fiign in the values of the lime ?
When only can the time occupied hy the traveller from A be negative ?
HowHhen will both e^Lpressions for the time occupied by the other
traveller be affected, with the positive or negative sign f
42. Same problem as the last^ except that the traveller from A is 9
days between the points B and G, and the traveller from G, 25 days
between the points B and A.
Ans. Time of meeting, db 15 days; tin^e occupied by traveller from
A, 24, or — 6 days ; time occupied by the other traveller, 40, or — 10
days.
43. In a certain bank there aro $438 worth of 5 and 3 cent pieces,
the number of the latter is exactly the square of the number of the
former. Required the number of pieces of each kind.
Am. 120 five cent pieces ; 14,400 three cent pieces.
44. Required to find two numbers, such that their sum, their pro-
duct, and the diflference of their squares, may all be equal to each other.
Ans,af^i + v/J, a/' = i — v'T; 3^ = 2 + V^i ^ = i — v/f
45. In the year 1706 the French made a descent upon Charleaton;
but '^ South Garolina," says Bancroft, " gloriously defended her territory,
and, with very little loss, repelled the invadors.^^ A certain nnmber
of the French were killed and wounded, and 100 were taken prisoners.
The number of killed and wounded was to the number of uninjured,
including the prisoners, as 1 to 3. And the squaro of the number that
escaped in safety from the expedition, was to the square of the number
killed and wounded, as 6i to 1. Required the number of invaders,
and the number of killed and wounded.
Ans. 800 invaders, and 200 killed and wounded.
Verification. If 200 were killed and wounded, then 600 were un-
injured, and 200 : 600 : : 1 : 3. And, since 100 wero taken prisoners,
500 escaped without harm from the expedition, and (500)' : (200)" : :
6t:l.
Why is the value connected with the negative sign of the radical
rejected ?
46. In the year 1842 South Garolina converted the citadel at
Charleston, and the magazine at Columbia, into military academies,
which were to be supported by the sum of money appropriated annuaBj
EQUATIONS OF THE SEOONB DEOBEE. 321
previous to this time to a goard of soldiers. The interest upon this
stun for 21 months^ amounted to $1960, and the square of the interest
for 6 months exceeded the square of 100*>' part of the sum by $288,000.
Required the sum appropriated annually to the military academies, and
the rate of interest. Ana, $16,000, sum ; 7 per cent, interest.
47. A man in Cincinnati purchased 10,000 pounds of bad pork, at
1 cent per pound, and paid so much per pound to put it through a
chemical process, by which it would appear sound, and then sold it at
an advanced price, clearing $450 by the fraud. The price at which he
sold the pork per pound, multiplied by the cost per pound of the
chemical process, was 3 cents. Required the price at which he sold it,
and the cost of the chemical process.
Arts, He sold it at 6 cents per pound, and the cost of the process
was i cent per pound.
48. The fore wheel of a wagon makes 12 more revolutions than the
hind wheel, in going 240 yards; but if the circumference of each
wheel be increased one yard, the fore wheel will make only 8 more
revolutions than the hind wheel, in the same space. Required the
circumference of each.
Ans. Circumference of fore wheel, 4 yards ; circumference of hind
wheel, 5 yards.
49. In the year 1853 there were a certain number of Woman's
Rights conventions held in the State of New York. If 6 be added to
tbo number and the square root of the sum be taken, the result will be
exactly equal to the number. Required the number. Ans. 3.
How does the negative solution arise ? Why is it neglected ?
50. A planter purchased a number of slaves for $36,000. If he had
purchased 20 more for the same sum, the average cost would have been
$150 less. Required the number of slaves, and their average price.
Ans, 60 slaves ; average price, $600.
51. A planter purchased a number of slaves for m dollars. If he
had received n more for the same sum, their average price would have
been c dollars less. Required the number of slaves and their average
price.
V
322 QENSBAL PROBLEMS INYOLVINa
' , n , /^nm + en* „ n /4nwi + «»*
fH . fK
x' and a/' expressing the number of slaves. Then, -^ and — , will ex-
press the average price.
Now, when n is zero, the two values of x ought both to be infinite,
indicating an absurdity. But the expressions will not point out an
absurdity unless reduced to their lowest terms by extracting &e
indicated roots. Then the two values of x may be written — — ±
(
n m m'^
-\ -J-, plus any other terms containing the higher powers
of X in the denominators. I
Now, make n ^ 0, and the two values of x both become infinite.
An expression can never be correctly interpreted unless it is reduced
to its lowest terms, for, as in the present instance, there may be a
common factor in all its terms, and a particular hypothesis made upon
that common factor may lead to absurd results.
What do the values become when c = ? what when n = and
52. The field of battle at Buena Vista is 6} miles from SaltiUo.
Two Indiana volunteers ran away from the field of battle at the same
time ; oHe ran half a mile per hour faster than the other, and reached
Saltillo 5 minutes and 54^^^ seconds sooner than the other. Required
their respective rates of travel. Ans. 6, and 5i miles per hour.
53. The New York shilling is 12} cents. A merchant bought a
quantity of cloth for $60. The number of shillings which he paid
per yard was to the number of yards he bought as 1 to 4y®Q. Required
the number of yards and the price per yard.
Ans. 48 yards ; 10 shillings per yard.
54. A grocer sold 1000 pounds of coffee and 1500 pounds of rice
for $240 ; but he sold 500 pounds more of rice for $80 than he sold
of coffee for $60. Required the price per pound of the coffee and rice.
Ans. Coffee, 12 cents per pound ; rice, 8 cents per pound.
55. A, B, and C, entered into partnership, and gained as much as
their joint fund, wanting $200. A's gain was $240. He invested
EQUATIONS or THC SECOND DEGBEE. 323
$100 more than B ; and the joint inyestment of B and C was $700.
Required the gain per cent., and the amount invested by each
partner.
Am, 80 per cent. A's capital, $300 ; B's, $200 ; C's $500.
56. There is a vessel containing 25 gallons of wine ; a certain quan-
tity is drawn out, and its place supplied by water. As much is drawn
out of the adulterated liquor as was first drawn out of the pure wine,
and there is now but 16 gallons of pure wine left in the vessel. Be-
quired the quantity of pure wine drawn out each time.
Ans. First draught, 5 gallons ; second draught, 4 gallons.
57. Same problem as last, except that the vessel contains ffi gallons
of wine, and that, after four draughts, — gallons of pure wine are left.
Ans. First draught, m — y/m ; second, = — ; third, ;
fourth, =-.
These results can readily be verified ; for, when the four draughts
are taken from m, there will remain
v/»» + = i 1 = — , which IS equal to v^m
^ -—1 1 — y/m 1 1 — y/m
-hi — y/m + -^= — 1 + — ==-, or -== + ^- — , or
x/m v^m* y/m ^
y/m + 1 — \/m 1
y or — .
The equation to be solved was really one of the fourth degree ; but,
owing to its peculiar form, it was readily reduced to an equation of the
second degree. Only one of the four values of the unknown quantity
have been used.
58. Same problem as last, except that the vessel contains 64 gallons
of wine, and that, after four draughts, the ^^^ of a gallon is left.
Ans, First draught, 56 gallons ; second, 7 ', third, || ; fourth, -^j^.
Verify these results by adding them together, and subtracting their
*ium from 64.
324 GENERAL PROBLEMS INVOLVING.
59. Two cotton mcrcliaiits rent a house for a certain sum, with the
understanding that each shall pay in proportion to the number of bales
of cotton he puts in the house. A puts in 500 bales, and B as many
bales as makes his proportion of pay amount to $200. B afterwards
puts in 500 more bales^ and then his proportion amounts to $225. Re-
quired the sum paid for house rent, and the number of bales first put
in by B. Ana. House rent, $300. B first put in 1000 bales.
60. A gentleman has two sums of money at interest, amounting to
$1150. The larger sum, being put out {^^ per cent, less advantsr
geously than the smaller, brings only the same amount of yearly inte-
rest. At the end of ten years, the smaller sum, added to its simple
interest for the whole period, is to the larger sum, added also to its
simple interest, as 11 is to 11}. Bequired the two sums, and the per
cent, on each.
Ans. $550 at 10 per cent., and $600 at 9^ per cent.
61. A gentleman bought a rectangular piece of land, giving $10 for
every yard in its perimeter. If the same quantity of ground had been
in a square shape, it would have cost him $20 less. And, if he had
bought a square piece of land, of the same perimeter as the rectan-
gle, it would have contained Qi square yards more. Bequired the
sides of the rectangle. -4ns. 4 and 9 yards.
62. A and B, together, invested $800 in a speculation. A's money
was employed 8 months, and B's 5 months. When they came to set-
tle, A's capital and profits amounted to $451; B's amounted to $375.
Bequired the capital of each, and their gain per cent.
Ans, A's, $440 -, B's, $360. . 10 per cent. gdn.
63. The seventh page of a treatise on Analytical Geometry has 11
more lines than the twentieth page of a treatise on Optics ; but the
seventh page of the Analytical Geometry has 4 letters less in each
line than there are lines in the twentieth page of the Optics, whilst the
twentieth page of the Optics has as many letters in each line as there
are lines in the seventh page of the Optics. The total number of let-
ters on both pages is 3542. Bequired the number of lines and letters
in each page.
Ans. Analytical Geometry, 46 lines, and 42 letters in each line;
Optics, 35 lines, and 46 letters in each line.
EQUATIONS OF THE SECOND DEOBEE. 325
r
64. Bequired to divide the quantity, a, into two such parts, that the
sum of the quotients, arising from dividing each part by the other,
shall be equal to m.
Ans. First part, _ + _^_^. or^-^^_j-^;
, . a a /m — 2 a ^ a /
second part, ^ - ^^^ jjj-^, or - + ^^
m — 2
m + 2'
Are there four independent values ? When only will these values
be real ? When will one always be negative ? When will the two
parts be equal ?
65. Required to divide 10 into two such parts, that the sum of the
quotients, arising from dividing each part by the other, shall be equal
to 2^fi . Ans, 7 and 3, or 3 and 7.
66. Required to divide the number 100 into two such parts, that the
sum of the quotients, arising from dividing each part by the other,
shall be equal to 2. Ans, 50 and 50.
67. Required to divide 15 into two such parts, that the sum of the
quotients, arising from dividing each part by the other, shall be equal
to 9^. A718. 13}, and 1}.
68. Four numbers are in a continued proportion, each number being
an exact number of times greater than that which precedes. The dif-
ference between the means is 8, and the difference between the ex-
tremes 28. Required the sum of the means, and the terms of the
proportion.
Arts, Sum of the means, 24. The numbers are 4, 8, 16 and 32.
In this example, let the unknown quantity be the sum of the means.
69. Same example as preceding, except making the difference of the
means, a, and the difference of the extremes, b,
/ JL I
Arts, Sum of the means + a Ky 7 ^ ; greater mean a +
\ / ; smaller, a \ / =■ 5 a; greater extreme -^ +
>\/ "'Vll-'°' ^-'"-^'\/-
^4a ' + b'{b — 3a)
5 — 3a
28
326 SQUATI0N8 OF THE SECOND DEGREE.
The negative values have been rejected.
When will these values become infinite ? When imaginary ?
70. There are two numbers, such that the square of the first added
to their product is equal to m, and the square of the second added to
their product is equal to n. What are the numbers ?
-4n». First, =fc — ; second,
71. Two numbers are to each other as m to n, and the square of the
first added to the product of its first power by m, is equal to the square
of the second, added to the product arising from multiplying its first
power by n. What are the numbers ?
An%, First, 0, or — m; second, 0, or — «.
72. The sum of the squares of two numbers diminished by twice
their product and by twice the first number, is equal to unity ; and the
sum of their squares added to the first power of the second number, is
equal to twice the product of the numbers. What are the numbers ?
Am, First, 0, or — \) second, — 1, or — \,
TRINOMIAL EQUATIONS.
353. A trinomial equation is one of the form, x*» + a:' =s y, involving
but one unknown quantity and three terms, and having the unknown
quantity in one term affected with an exponent double of that with
which it is affected in the other. A trinomial equation then contains
two terms, in which the unknown quantity enters, and an absolute
term.
Let it be required to solve the equation, x^ '\' ^ ^:i 20.
Let a' =s y. Then the equation becomes ^ + y = 20.
The last equation gives y = + 4 and y = — 5. But, x* = y.
Hence, jc« = 4, o r — 5. Then, a/ = + 2, rr" « — 2, a/" == + V 6^
cc""s= — ^ — 6. Either of the four values will satisfy the given
equation, a^ + x' = 20. The substitution of the last value gives
(— V'^^* + (— sl^^^7 = 20, or, 25 — 5 = 20, a true equation.
Solve the equation, 7^ — a;' = 702. ^Make o^ z=: y. Then the equa-
tion becomes y* — y = 702. From which, y = -f 27, or, — 26. But
^ = y. Hence, x = t/27"= 3, and x = ^--26.
There are really four more values for a;, since, as will be shown here-
after, the number of values is exactly equal to the degree of the cqua-
SQUATI0N8 OF THB BEOONB BBQBSB. 827
tioD. But the method of detennining the other values belongs pro-
perly to the general theoiy of equations. The two values found can
be readily verified.
The examples given are of a simple character. The equations were
already of the proposed form. But it frequently happens that an arti-
fice must be employed to put the equation under the form of x*' + x"
=s q. Take, as an example, x" -f- v^x* -h 9 =» 21, Add 9 to both
member s, and w e have x* + 9 + ^/x* -f 9 = 30. Make x* -f 9 = y",
then, v^x* + 9 = y, and the equation becomes ^ -f y = 30. From
which we get, y -^ + 5, or — 6. Then, x' + = 25, or, x« + 9 = 36.
The values of x are -|- 4, — 4, -h >/27, and — %/27. The negative
sign of the radical, ^x^ + 9, must be taken in connection with the last
two values. This is indicated by the equation, v^x* + 9 ^ y = — 6.
Again, take Of!±l)' + i = fl±_? + 91.
° X X
This may be wntten, ^^ ■— — ^^ = 90.
•^ ' X X
Let = y. The equation then becomes t/* — y = 90. Hence,
X
x'-l- 9 x'4- 9
y = 4- 10, or — 9. Then, —^— =10, and — ^— « — 9. These
X X
9 9
equations gives the four values, + 9, + 1, and — "o" + J V^, — -5-
— is/l^
The equation, t/a? + 1 + t/x* + 2x + 1 = 20, may be changed
into ^x+ 1+ V^ + 1 = 20. Let Vx~+1 = y. Then ^x + 1
=s y", and the equation becomes t/" + y = 20. From which, y as 4, or,
— 5. Then ^/x -f- 1 = 16, and ^x -f 1 = 25. Hence, x = 255,
or, X = 624.
The equation, x' + ^1? + 56 = 34, may be placed under the pro-
posed form by adding 56 to both members. Then, x* -[-56+>/x* H- 56
ras 90. Let >/x* -t- 56 = y. The equation then becomes y" + y = 90.
From which, y = -f 9, or, — 10. Then x = -f 5, — 5, + \/44,
— -v/fiT
These illustrations are sufficient to explain the spirit of the process.
No general rule can be given. Any modification may be made upon
the equation that will place it under the proposed form.
328 EQUATIONS OF THE SECOND DBQREE.
GENERAL EXAMPLES.
1. Solve the equation, x| + x} = 12.
Ans. x= + 27, or — 64.
2. Solve the equation, x -{■ zi i= G. Ana. x as 4, or 9.
3. Solve the three equations,
2" + zy + y* « 1900 (A).
a;« + a:z + z" = 1300 (B).
f + xy+a^^ 700(C).
Subtracting (B) from (A), and (C) from (B), and factoring, we get,
600 ,^^ J . 600 ,^^
y + 2 + a: = (D), and y + 2 + a; = (E).
y — ^ z — y
X -^ z
By equating (D) and (E), we get y = — - — . This value of y, sub-
stituted in (D), gives z-^- x = • , or, z^ — a;*=800; from which 2* =
Z ■'^■~ X
a;' + 800. Substituting for z its value in B, we get a:" + a: y/x^ + 800
-f x» + 800 = 1300. From which, x v'x»'+~800 = 500 — 2jc*.
Squaring and reducing, there results, 3x* — 2800a;' = — 250000, or,
2800x« 250000
X
3 3 •
The combination has led to a trinomial equation, which, when solved,
will give, for one system of values, x = 10, y = 20, and z = 30.
4. Solve the equations,
y + ary + ic" = 7,
y* — a:y — X* = — 6.
Ans, X = + 2, — 3, y SB +1 — 1.
5. Solve the equations,
X + y + » = 6,
x» + y» + «" = 14,
XJ8 s= 3.
Arts, y = 2, X as 1, « = 8.
One system of values only given.
6. Solve the equations,
y» + x" « 2600,
y« — 2y = 2x. '
Ans, y =s 10, and x = 40.
One set of values only given.
EQUATIONS OF THE SECOND DEOBEE. 320
7. Solve the equations, y*+a:" = y" + y* + 8,
y» — 2y=2a:.
Ans. One set of values, y =^^) ^^^ a; s 6.
8. Solve the equation, ^a^ + 4+ ^x* + 45=6.
Ans. One set of values, x' = + ^12, as" =s — ^12.
9. Solve the equation, s/}^—y 4- y/r^—y =3^10 + ^90.
Ans, One set of values, y = 10, or y =» — 9.
The positive value of the radical only is taken.
10. Solve the equation, a;* + 7 + y/a^ -|- 7 =s 20.
Ans. a/ = + 3, a/' = — 3, a/" = + v^; «*^ = — v/lS^
PROBLEM OF THE LIGHTS.
354. Two lights are placed on the same indefinite right line, the one
shining with an intensity represented by a, the other with an intensity
represented by h ; the problem is to find, on this indefinite line, the
point or points of equal illumination — assuming a principle of optics,
that the intensity of a light varies inversely with the square of the
distance from that light.
A IB
— j> J J —
Let A be the position of the first light ; B, that of the second.
Let m =B AB, the distance between the lights.
Let a SB intensity of first light, at one foot from A.
Let h as intensity of second light at one foot from B.
Let I be the unknown point of equal illumination.
Let B I ^ m — x =s distance of same point from B.
Now, in accordance with the assumed optical principle, the intensity
of the first light, two feet from A, will be expressed by x y'^r 2" : 1" : :
a : -j-/, at three feet, by -^, and at x feet, by -^, and this last expres-
sion wHl represent the intensity of the first light at the unknown point,
I. In like manner, the intensity of the second light, at B, will be
28*
830 EQUATIONS OF THS SECOND DEGREE.
denoted by r^. But, by the oonditions of the problem, the in-
tensities of the two lights must be equal at I. Hence, we have — ^ =:
r^, which may be changed into r^ =* -=-. and, by eztnct-
(7n — xy ^ ® (m — x)* 6' ' -^
ing tlve root, we get = zt — =. We will, for conyenience, call
x' that value of x which is connected with the positive sign in the
second member, and wc wiU call a^' that which is connected with the
negative sign. Hence, -z =s ^^, and -j, = =. And
m — ar ^i m — x ^i
solving these equations, we get a/ = — =z =^, and x" = — z=r
Now, since there are two distinct values for x, we conclude that in
general there will be two points of equal illumination. By this we do
not mean that there will be two points of equal brilliancy, but that there
will be two points where the intensities of the two lights will be equal
to each other. By subtracting oi and x" in succession from m, we get
tn — X !^ — zzz z:, and tn — x =s — — = =..
y/a + y/b y/a — v/6
Hence, we have the system of values:
a/ = — =, distance from A to first point of equal illumination.
y/a + y/h
m-a^= '^^ _ , distance from B to same point.
a/' = -^= =y distance from A to second point of equal iUomina-
y/a y/a ^^^^ j£ there be one.
m — x" = =: — ^r=, distance from B to same point.
y/a — ^/h
Now, these expressions have been deduced upon the supposition that
the point, I, was between A and B, and, consequently, we have assumed
that the distance A I is positive, when estimated on the right of A,
and the distance B I positive, when estimated on the left of B. Hence,
if either a/ or re" becomes negative in consequence of any imposed
EQUATIONS OF THE SECOND DEGREE. 881
condition, the corresponding point of equal illamination will be found
on the left of A. And, in like manner, if either m — sf or m — of
becomes negative, the point will be found on the right of B.
We will begin the discussion by supposing a ss b. Then, of s
m^a m>Ja m AB_ -x^ri^
— = m = — =1 = i7-> or A I = ~7r-' Hence, the point is half
y/a+y/b 2>/a ^ 2 > i-
way between the lights, as it obviously ought to be. Now, m — a/,
which expresses the distance from B to I, ought to give the same point.
It does so, for f» — ar = -^= =^ = — = = tt-. We wul next
examine whether there can be a second equally illuminated point, when
the intensities of the two lights are the same. We have a/' = — ^r—
^ 00, and m — cc" = tt— = — oo. These values indicate that
the second point is at an infinite distance on the right of B ; or, in
other words, that there is no such point at all. This ought to be so,
for, when the two lights are of equal intensity, the one cannot throw its
beams with the same power to a greater distance than the other, and
this must be the case if there be a second point. The second values
(jxf* and m — a/') then denote impossibility. By going back to the
equation of the problem, we see that it will become, when a ^ 6, o^ ss
(m — x)', an equation which can only be true when m = 0, or a: = -^.
But m s= is contrary to the hypothesis ; hence, a; s --• is the only
true value. The assumption then, that there was a second point (a
being equal to 6), was absurd, and led to oo , the appropriate symbol of
absurdity.
We will next suppose a^b.
Then, a/ = — = becomes *>. -s-, because, if the denominator
s/a+s/b 2'
m
were 2 v^a, the value of the fraction would be ~2~; ^^^ since the deno-
m
minator is less than 2v^a, the value is greater than ~2 *
Hence, the point, I, is nearer B than A. In this case,
A IB r
882 EQUATIONS OF THE SECOND BEQBEE.
m — a/ <[ 7p, for, were the denominator of the fraction ^ — — :;:_
m ^
exactly 2y/b, the value of the fraction would be "9"; but since the deno-
minator is greater than 2 ^o, the value is less than -^. This result
corresponds to the former^ and places I nearer 3 than A ; that is,
nearer the light of feebler intensity, as it ought to do. We see that s^
is greater than m, for were the denominator of a/', the ^a, its value
would be equal to «i, but as ^/a — >/b ia less than y/a, vd' \a greater
than m. Hence, the second point of equal illumination is beyond B.
In this case, m — a/' = ^= =:. becomes negative, since the
v^a — y/b
numerator is negative and denominator positive, and, therefore, the
second point must be on the right of B. The results then agree with
each other, and agree with the fact. For, after passing the point I, the
intensity of the 1st light becomes feebler than that of the 2d. At the
point B, there is the greatest difference between their intensities.
Beyond B, both lights diminish in brilliancy, but the 2d more rapidly
than the 1st, because of its greater feebleness; and we at length reach a
point, r, where the illumination is equal.
Next, suppose b^ a.
Then, a/ := — = ^ <* — . For, were the denominator 2 s/a, Ac
value of the fraction would be -^, but as the denominator is ^ 2 yfa,
the value of the fraction would be <^ -5-
FA I B
Hence, the point, I, will be found nearer A than B. We have m — x' ^
TO \/6~ ^^
— > "2 ; for, were the denominator 2 v^S^the value would
y/a + ^b
be = --, but as the denominator is <^ 2 y/Uy the value is ^ — . The
results then agree, and place the point, I, nearer the feebler light
Again, we have a/' =3 _ __ affected with the , sign minus, the
y/a — 4i/b
EQUATIONS OF THE SECOND DEGREE. 833
numerator being positiye, and denominator negative. The second point.
Vy most then be on the left of A. By multiplying the numerator and
denominator^ of the value of m — a/', by minus unity, we will have m —
a/' =5 — = > w ; for, were the denominator = ^~F, the value
would be equal to m, and, since the denominator is <[^ ^6, the value is
greater than m. Hence, the second point is beyond A at 1", as has
just been shown. Were the values of m, a, and &, given, the position
of the point of equal illumination could be readily determined. Let
i» = 20 feet, a = 16, 6 = 36. Then a/ = 8 feet, and m — a/ = 12
feet, and the intensity of the two lights will be equal at the point cor-
responding to these distances. For, the intensity of the 1st will be
— 5 = J, and that of the 2d 7^2)8 = *
r A I B
It can readily be shown that, between I and A, the intensity of the
Ist light is greater than that of the 2d. Thus, at 2 feet from A, that
of the 1st light will be expressed by 4, and that of the 2d by ^. At
the point I", 40 feet on the left of A, and 60 feet from B, the intensi-
ties of both lights will be expressed by j J^th, and are, therefore, equal.
And, by assuming other points, and calculating the corresponding
intensities, we would find that they were only equal at I and I".
Now, let m = 0, and a unequal to h, then x', m — x', x", and
m — x" = 0. These are plainly absurd solutions, for the two lights,
shining with unequal brilliancy, cannot, equally, illumine the point
at which both are placed. By recurring to the equation of the problem.
x' a
it becomes, when m =0, — ^ = -7- This equation can only be true
when a = b, which is contrary to hypothesis.
Now, let m ^ 0, and a ^ b.
Then, x' = 0, m — x' = ; x" = -jr-> and m — x" = -rr-. The first
two values (x' and m — x',) show that the point at which the lights are
placed, is equally illuminated. The second two (x" and m — x",) arc
the symbols of indetermination. This ought to be so, for when the
equal lights are placed together, the points, at one foot, two feet, three
feet, &c., are as much irradiated by one light as the other. There is^
384 EQUATIONS OF THE SECOND DEGREE.
therefore, no particular .determinate point at which the intensities ure
equal, and we say that the problem is indeterminate. By reourring to
the equation of the problem, ; = = -p, we see that when m = 0,
^ ^ (w — xy b
and a = 5, it reduces to an identical equal equation, x' s= a?, which can
be satisfied for any values of x.
Whenever we get ^ as a solution to a problem, we can tell, by recur-
ring to the original equation, whether it indicates in determination. If
the particular hypothesis, which reduces the solution to -rr-, also reduces
the original equation to an identical form, we have a case of indeter-
mination. But, if the given equation do not reduce to an identity, we
have a vanishing fraction.
Another test, also, may be employed. If the value becomes -jr in con-
sequence of a single hypothesis, we know, certainly, that we have a
vanishing fraction, and not an indeterminate solution. The^ preceding
values become -^ in consequence of the two hypotheses, m =b 0, and
a = h, and might or might not be vanishing fractions. Thus, tlie ex-
(^ a*)(x h) . . . " .
pression, ^ \ A^ i tjy ^ * double vanishing fraction for a: = a, and
\X ""— d) {XT J
x=zb.
Now, suppose the first light extinguished ; then, a a ; and we
find a/ and x" == 0, and m — x', and m — x" = f». These solutions,
at first, seem absurd, since they indicate that the point of equal illumi-
r A I B
nation is at A. But, suppose the light at A to be very feeble, there
will then be two points of equal illumination, I and T, very near to
and on opposite sides of A. Make the first light stUl mor« feeble, and
the two points, I and I', will approach nearer to A ; and, finally, when
that light is extinguished, they will unite at A.
Next, suppose both lights extinguished, or a sa 0, and 6 s 0.
Then, x', x", m — x', and m — x" become -jr-, indeterminate, as they
ought to be.
Suppose m SB 00, or that the lights are infinitely distant from eftcb
T7NDETEBMINEI) COEFFICIENTS. 885
other. ThBJif if a and h are finite, the values all become infinite, as they
ought, since there can be no point of equal illumination. The equa-
ct h
tion of the problem, — = = -, can be satisfied, when w = oc,
or (m — xy '
by making a; = 0, which places the point of equal illumination at A;
or, by making x = m, which places the point of equal illumination
atB.
Suppose, a =ss 00 ; then, a/ and a/' both become equal to «i, since
^b may be neglected in the denominators of the fractions, and the
point of equal illumination is then at B. The values of m — a/ and
tn — x" become zero, and indicate the same point. This ought to be
80 ; for, by making the intensity of one light infinitely great, we make
that of the other relatively infinitely small, and we then ought to get
the same result as we did when one of the lights was supposed to be
extinguished.
UNDETERMINED COEFFICIENTS.
355. The method of undetermined coefficients is used to develop
algebraic fractions into a series, and to determine the value of the con-
stants which enter into identical equations ; that is, equations which can
be satisfied by any value whatever, attributed to the unknown quantity.
For the development of fractions we assume the form of develop-
ment ; and we are governed in our assumption of this form by our
knowledge of what it ought to be. Thus, if we proceed to expand
ft fit 3J
by the ordinary process of division, we obtain — ■— = 1
a+x '' ^ a+x a
jp* X* X* X*
H — 5 iH — i k+i&c. The quotient is an infinite series,
or a^ a* a^ ' ^
arranged according to the ascending powers of x, and having x in each
term affected with a positive and an entire exponent. The first term may
be supposed to contain x, affected with a zero exponent, and the coeffi-
cients of X are then 1 1 — 5 i + &c.< and it is evident that
at a' a*
these coefficients are entirely independent of the particular values that
1 1 1 , .
may be attributed to x. They are stiU 1, 1 — , — ^ "^ *^''
886 UNDETERMINED COEFFICIENTS.
when a; ^ 0, 1, 2, 1000, anything whatever. We may remark, in
this connection, that the independence of the coefficients upon the
value of the letter with which they are associated is not confined to the
expansion of fractions, hut is true in all developments whatever. Thus,
(a 4- xy =s a* + 2ax ■\- a? ; the coefficients are a", 2a, and 1, and
these coefficients will not he altered hy any change in the value of x.
We have seen, by performing the division, that the exponents of jt,
in the development of , must be positive and entire, and that
the coefficients must be independent of x. If, then, we were required
to assume the form of development of this fraction, or any like it, we
would know that the assumed form must fulfil the required conditions.
Now, since a may be regarded as the coeffi^cient of x^^ the fractioni
, may be considered as arranged with reference to rr, both in the
numerator and denominator; and we have seen that a fraction so ar-
ranged, beginning with the zero power of x, gives positive and entire ex-
ponents for X in the quotient. This law can readily be shown to be gene-
ral, and it is not even necessary that the zero power of x should enter
into the numerator. If the exponents of x in the numerator are all po-
sitive and entire, and the terms arranged according to the ascending
powers of x, and if the denominator is also arranged in like manner, and
contains a constant, a\ that is, a term involving oc^, then all the exponents
will be necessarily positive and entire. For, take the general fraction,
-^ — ^-f — .-^T— = A + Bx »" + Car' -h &c., (N), in which the second
a H- ox^ -f cx^
member represents the development of the fraction. Now, since a frac-
tion and its development must constitute an identical equation, it is
plain that, from the nature of such equations, (N) must be true when
a: = 0. But Ca;~', which represents the term (if any) containing a
negative exponent, can be written — , and is equal to infinity when
X
X = 0. Hence, the whole second member is infinite. But, x = ;
reduce the first member to — 7 ; n, m, />, and q being supposed posi-
(JL
tive. Now, if a = 0, we have — j = = second member = 00, which
is absurd. If a be not zero, still we have -7 = 00, or a finite quantity
equal to an infinite, which is absurd.
UNDBTXEMINBB COSFTIOIENTS. 387
Jtf iken, all the exponents in the numerator and denominator are
positive and entire^ and the denominator contains a constant^ or both
numerator and denominator contain constants, all the exponents of the
development must be positive. We will now show that all the expo-
nents of x must also be entire. We have represented, bj Bx"^, the
term, if anj, involving a fractional exponent. Now, suppose — sa i.
Then, Bx " = Bx = B^/x. But every square root has two distinct
values. Hence, Bv^x has two distinct values, and the whole second
member has two distinct values. But no fraction containing only entire
exponents can give two quotients. The assumption, then, of a frac-
tional exponent in the development is absurd. If — = ij !> |> &c., the
fHf
m
term Bx"^ would have three values and three quotients, which is ab-
surd. So, if — = any fraction, there would be more than one quotient.
m
We are now prepared to assume the form of development of any al-
gebraic fraction, arranged according to the powers of a certain letter,
and having that letter in the first term of the numerator affected with
an exponent equal to, or greater than that of this letter in the first or
last term of the denominator. The exponents of the arranged letter
are also assumed to be positive and entire, both in the numerator and
denominator.
Let us then place — : — = A -f Bx + Cx* + Dx* -f Ex* + &c., in
which A, B, C, D, &c., are independent of x, and in which all the ex-
ponents of X are positive and entire. A is the representative of that
term which does not contain x, or contains x affected with a zero expo-
nent. There is generally such a term in a development, and there
must, of course, be a representative of that term. Clearing of frac-
tions and arranging according to the ascending powers of x, we have
a aa Aa + << B
+A
X + aC
+B
x^ + aD
+ C
+ C
x^&c. (M).
Now, since the coefficients are supposed to be independent of x, their
values will not be affected by making x = 0. If then we can deter-
mine these values when x = 0, they will be true when x = 1, 2, 3 ;
anything whatever. Making x = 0, we have a = Aa. Hence, A ^ 1.
29 w
388
UNDETERMINBD COBPEIOIBNTB.
Now, since a « Aa, these two terms cancel each other, and (M) becomes
X + aC
x' + aE
+D
X* + &c.
= aB
+ A
Dividing both members of this equation by x, which we have a light
to do, we get
= aB + aC
+A +B
X + al)
+C
+D
a? + &c. (N).
Making, again, x = 0, we have left, = aB + A. Hence, B = —
:^ -_ Jl. Now, since aB + A has been found to be zero, they may
a a
be stricken out of (N), and that equation will bec9me
0= aQ
+B
x + aD
x'+ aE
x' -f &c.
And again, dividing out by x, we have
0= aC H-aD
+B +C
x" + aE
+ 1)
x« + &c. (P).
Making x = in (P), we have left, = a + B. Hence, C =
"B 1
— Z- = -j — -. Omitting the term, aC + B = 0, in equation (P),
a or
and dividing again by x, we have
= oD + aE lx + &c. (2).
+0 +D 1
Making x = 0, we have left, = aD + C. Hence, D = =
-, Dividing (2) by x, and again making x = 0, we have aE + D
— ■■ + -J. Now, if we substitute for A, B,
a a
a*
= 0. Hence, E =
a
C, &c., their found values, in the original equation, — ^ — = A + Br
a "T" X
4- Cx" 4- Dx" -f Ex* + &c., we will have
a - X x^
a + X a a"
X* a^ x^
-. + r + &c. ; the same result that we obtained by division
By transposing a to the second member in equation (M), we have
= Aa
xP+ aB
+ A
X + aC
+B
+C
X* + &c.
UNDETERMINED COEFFICIENTS. 339
And, since we have found Aa — a = 0, a A -f- A = 0, aC + B = 0,
aD -f C = 0, &c., we conclade, that, if we have an equation whose first
member is zero, and whose second member contains all its terms ar-
ranged according to the ascending powers of a certain tetter ^ the expo-
nents of this Utter being all positive and entire^ and its coefficients in-
dependent of it, these coefficients will be separately equal to zero.
This is the first enunciation of the principle of undetermined co-
efficients, and ought to be remembered.
The second enunciation, (which is an immediate consequence of the
first), is as follows : if we have an equation of the form, a -h bx + co(^ •\'-
da? + «a;* + &c., = a' + b'x + <fa? + ^^ + ^^*^ + &c., which is satis-
fied for any value of ar, then the coefficients of the like powers of x
in the two members will be respectively equal to each other.
For, by transposition, we have = (a' — a) x^+ (1/ — b) x +
(</ — c) a?'\' (cP — d) a?-\' (e' — «) x* -f Ac. And, since these coeffi-
cients are independent of x, we must have, by what has just been shown,
a' — a = 0, 2/ — 6=0, </ — c=0, d' — e? = 0, c' — e = 0, &c.
HencC; a' = a, b = bf cf z= c, d = d, ef =. e, sa enunciated.
The preceding equation, a + bx + cx^ + da? + ex* + &c. = a' -f-
b'x + (/a? + &c., and all other equations to which the second enuncia-
tion is appliablc; belong to a class of equations called identical equa-
tionS; the two members of which differ only in form.
It is, in general, most convenient to develop expressions in accord-
ance with the second enunciation.
Bequired the development of ^ .
1 X
Let r-- — = A + Bx -f Cx« + Daj' + Ex* -f- &c.
1 -f- X
Clearing of fractions, we have 1 — x = A-f(B+A)x-|-
(C + B)x« -I- (C + D)x8 H- (E + D)x* + &c. By placing x = 0, we
have A = 1, and we might proceed as we did with the fraction, ,
continually dividing by x, and making x = in the resulting equation.
But, by the second enunciation, we have at once A = 1,B+A = — 1,
C + B = 0, C + D = 0, andE-fD = 0. From which, A = 1,
B= — 2, C = + 2, D = — 2, E = — 2. And, substituting these
1 X
values of A, B, C, D, E, &c. in the equation, «: A + Bx + Cx*
1 H-x
1 X
-f-*c., we get T =1 — 2x + 2x'^ — 2x« -f 2x* — &c.
1 -|- X
340 UNDBTEEMINBD COEFFICIENTS.
This is the same result that we would obtain by actually performing
the indicated division.
We have assumed the form of development to be A -f Ba? + Cx* + Ac.,
in which the exponents are all positive and entire. Now, if we have
an expression whose development must necessarily contain negative or
fractional exponents, it would be absurd to place it equal to A + B + x +
Cx' 4- &c., and the result will make manifest the absurdity by the
symbol, oo . Suppose it be required to develop ;^. It is plain
that the first term of the development is x"' ; if, then, we attempt to
develop the expression by the method of undetermined coefficients, we
commit an absurdity, and that absurdity ought to be made manifest io
the result by the appropriate symbol, oo .
Let, then, — ?— . = A +Bx + Cx« + Dx» + Ex* + &c.
X — x'
From this we get l==Ax+(B— A)x»+(C— B)x»+(D— C)x*+Ac
Now, in accordance with the first enunciation, make x = 0, and we
get 1 = 0, an absurd result. But, if we first divide both members by
X, and then make x = 0, we will have J z=: c» = A, and the absurdity
is pointed out by its appropriate symbol.
The above expression can be developed by changing its form. Do-
1 . . 11 1
compose -j into its factors, — X z , then develop := by
X "^~ Xr X X — ~ X i — — X
the method of undetermined coefficients, and multiply the development
by — . By developing, we find = = 1 + x-\'a^ + a^ -f x*4- &c. ;
X A ""•- X
111 , «
consequently, -^ = — X .= — ; — =:x-*+x*'4.x + a^+ x* +
X """ JU X JL ^"^^ X
X* + &c. The series is evidently arranged according to the ascending
powers of x. For most expressions, a slight inspection will show whe^
ther their development will contain negative or fractional exponents.
and, consequently, whether the assumed fprm is right. Thus, = ^
1 — X ■
cannot be placed equal to A + Bx + Cx', &c., because its developmeni
must contain negative exponents ; j cannot be placed equal to
1 +x^
the assumed form, because its development must contain fractional ^-
ponents These expressions, and others like them, can, however, be
UNDETBBMINID 00EFFICIENT8. 341
expanded into a series by ohanging their form. Pkce ar* = z, then^
11 •
1— X-*"" l — z
Let 1 — a? = A + Bz + 02* + Dz» + Ez* + &c.
1
We will find = = l + « + 2« + ;5» + 2* + &c
1 — z
Now, replace z by its value, and we have = — » = 1 + xr^ + xr^
In like manner, place x^ = z. Then, .- = = = A -\-Bz
+ 02;' + Dz* + E5J* + &c. We will find by the method of undetermined
1
coefficients, ^ =1 — z +z* — 2*+ 2^, &c. ; and, replacing z by its
1+2
1 14
value, we have = 1 — x'^ + a; — x* + a* + &<5.
1 + 0-2
FAILING CASES.
356. Any fraction, all of whose exponents are positive and entire in
the letter according to which the series is to be arranged, can be de-
veloped, provided the denominator contains a constant. Again, any
fraction, all of whose exponents are positive and entire in the letter
according to which the series is to be arranged, can be developed ; pro-
vided, that there is one term, at least, in the denominator, the expo-
nent of whose arranged letter is equal to or less than the exponent of
the same letter in, at least, one term of the numerator. A fraction
that fulfils either of the foregoing conditions can always be developed,
and the development will be the same as the quotient. The reason of
this is plain ; the method of undetermined coefficients gives a develop-
ment for fractions, which is the same that would be obtained by actual
division, beginning with the lowest power of the arranged letter. Now,
a fraction that fulfils either of the foregoing conditions, will, when ex-
panded by division, necessarily give positive and entire exponents in
the arranged letter, when the division has been begun, with the nume-
rator and denominator arranged according to the lowest power of this
letter.
Division would give us two quotients for — ■— y, according as we
a; + 1
made a;, or 1, the first term of the divisor; but the method of undeter-
29*
842 UNDETEBMINSD C0EFEI0IENT8.
mined coefficients gives but one development, which is the same as
that obtained by arranging the numerator and denominator with refe-
rence to the lowest power of x. Therefore, the formula, which is ap-
plicable only to positive and entire exponents, must fail, when the least
exponent of x in the numerator is less than the least exponent of x in
x' -i_ a^
the denominator. Thus, -y— — cannot be developed, because a', which
X "f" X
involves the lowest power of x in the numerator, gives a quotient
affected with a negative exponent when divided by x, the correspond-
ing term of the denominator.
GENERAL EXAMPLES.
1. Develop 1 + x. Ans. 1 -^ x.
2. Develop =-— , into a series.
^ 1 + X 4- cc*
Ans, 1 — x + x^ — cc* + xf — x' -{- x^ — &c.
1 5a;
3. Develop = j- into a series.
Ans. 1—x — ^a^—lQa^ — Ux* — 256a;» — 1024a:« — &c.
1 ^
4. Develop -z into a series. . , -
1 + X Ans. 1 — X.
5. Develop = z^ into a series.
X ~~~ X
An*. x-^ + x-' + ar* + x-* + a;-" + &a
6. Develop j into a series.
l — x^
I % 4 f
Ans. a:' -h x' + X + a' -f- a;' + a:" + &c.
1 + X .
7. Develop ^ ^ into a series.
Ans. 1 + 3x + 6x2 + 12^^ + 24x* + 48x» + &c.
8. Develop = — ; 5— — s ^^^ ^ series.
1 -f- X + X -f- X
Ans. 1 — x + x* — x' + x' — x* + a?" — x** +Aa
TTNDETEBMINED G0EFFICISNT8. 343
9. Develop __-^___ into a serieaf
Ans, 1 — aj + a:* — a;«+a»« — aj"+x»* — x*+a:» — x«» + &o.
10. Develop -^. Am. — --7.
11. Develop ■ into a series.
Ant, 33* — x' + aj* — 05* + aj* — x^ -f &c.
a;* + a;'
12. Develop -—=- into a series.
X -J- o
. Q^ , 40;* 4a;* 4a;* -
^"*--6 +25■-i25 + 625-*'•
^■— y :.
13. Develop ^ into a series arranged according to the powers
X y
of y. An9. a?--' + x^-^y + x»-y + af^y^ + y».
x' + a;*
14. Develop
^ •\- x' Am, X.
iR Ti 1 a;» + 3x« + 3x + l . ^
10. Develop -— ! — mto a senes.
X T* X
-4»«. x* + 2x + 1.
x*
16. Develop , into a series.
X "^ X
.4««. x' — x' + x* — x' + x' — x' + &c.
-H, ^ 1 x^ + x* + x' .
17. develop J-- into a senes.
X -7- X
^ri«. X + x* — x* + x« — x^ + x' — X« + &0.
18. Develop -^ — ^i— mto a series.
XT ^~ X
-4»w. l+ic" — ar» + 2x* — 2x* + 2x« — 2x^ + 2x» — 2x»+&c.
^^- ^'^'^^P x + x^ + S + x^ ^"^^^ ^ ''"^- -
Am, 1 — x« + x» — X* + 2x« — 3x^ + 2x + x" — 5x"*.
on -n 1 a;' + Sx* + 3x + 1 .
20. Develop j-II Z_ mto a series.
X + X Am, 00.
The process fails. Why ?
844 UNDETERMINSD COXFriCIZNTS.
1 J- 7ic + ic* .
21. Develop — "L into a series.
^ 1 + re
Ans. 1 -fdar — 6x»-f 5a:* — 5x* + 5x*— 5x« + Ac.
ar» — a«
22. Develop
cc — a' Ans. x -{- a.
23. Develop — — — , into a series.
a a or a
or* + x"" + 2 .
24. Develop _, ^ into a series.
Ans, 2 — a:-« + 2j^ — 2ar« + 2a;-* — 2ar » + &c.
Make ar* = z. Develop, and replace z hy ar^.
Remarks.
357. It will be observed that, in most of the preceding examples, any
term of the series after a certain number from the left, can be formed
from the term immediately preceding, or from two or more preceding
terms, according to some fixed law. Thus, in Example 2, every alter-
nate term of the series is formed from that which precedes it, by multi-
plying by — X, The series in 5 and 6 are formed in the same way, the
multipliers, or all the terms, being x~^ and x\. In 7, every term after
the second is formed from that which precedes it, by multiplying by 2x.
In 8, we may regard the constant multiplier as — x, and omit eveiy
alternate set of two terms. In 9, the multiplier may be regarded as
— Xf and, when three terms occur together, these are omitted. In 11
and 16, the multiplier is — x. In 12, the multiplier, for all terms after
the second, is — \x. In 13, the multiplier is x~'y. In 17, the con-
stant multiplier, after the second term, is — x. In 18, the multiplier,
after the fourth term, is — x. In 19, there is no law of fbnnatioD.
When the terms of a development are formed from those which pre-
cede according to some fixed law, the development is called a recurring
series. Multiplication is not the only mode of forming the succeeding
terms ; sometimes they are formed by addition or subtraction, and even
by a combination of two of these methods. Thus, = s
^ ' 1 — x — X* — X*,
expanded, becomes 1 + x -^ 2a;" + 4x* + 7a^ + ISx" + 24jt:* -h &c.
VNDETJBEMINSD OOBFVIOIENTS. 845
The litenl parts $xe formed by multiplying by x, the fiist two terms have
the same coefficient^ and each succeeding coefficient is equal to the sum of
the three which precede it, &c. Again, = ^ « 1 4. 8a; -f 4x'
1 — X — ar
+ 7x» + 11a;* 4. 18a:* + 29x« + 47a:' + &c., in which the coeffi-
cients after the second are equal to the sum of the two preceding
ooefficients.
PARTICULAB CASES.
358. A fraction involving irrational monomials may be treated as in
the following example.
Required, the development of rj ^. Let x = s^. Then, xi =»
X ^^ a5g
1^, a;| = j)^, 2 as xi. Then, ihe fiaction becomes = -^, and, by tlie
method of undetermined coefficients, -= -^ = A + B2; + Gs? + ^^
+ E«* + Fsj* + &c. From which we get A = 0, B = 0, C == 0,
D = 1, E = — 1, F = 0, G = 1, H = — 1, &c. Hence, f^
= «* — «*+«;• — ^ -{-7? — »*° + &c, and, by replacing z by its value,
we have 1 =xi — | + x — a;J + x| — x| + &c.
Since all examples of the same kind may be treated in t^ same way,
we derive the
RULE.
Place the variable equal to a new variable, affected with an «rpa-
nent equal to ike lea^t common multijple of the denominators of the
exponents of the old variables. Develop the new fractvm^ and replace
Ae new variable by its value in terms of the old.
EXAMPLES.
xj X?
1. Expand ^ ^ into a series.
Ans. X J — »f + icj — x| + &o.
X^ "" xi
2. Expand -y— — =. Ans, xi — a; J + x J — x\ + &c.
846 UNDETEBHINED COEFFICIENTS.
859. Fractions involviDg parentheses, differing only in their expo-
nents, may be treated as the following.
X
Required, the development of — -. , ,- .
^ ^ (1 4- a:)i — (1 + a?)
Letl + a: = ^. Then, ^^ ^^^^^^^ ^^^ ==^ which, when
developed, gives « + «? + «* + 2* + &c. Now, replace z by its valney
and we have ^^^^^^1^^^^^ = (1 + x^i + (1 + x^ +
(1 + x)| + (1 + xy + &c.
BULE.
Place the common parentkesis equal to a new variable^ raited to a
power denoted by the leaM common mvUiple of ike denominators of ike
parentheseSf and proceed as before.
EXAMPLES.
Ans. (1— x)J+(l — x)i + (l — x)« + (l — x) J+Ae.
2. Expand 7= ri — =-7= r^ into a series.
(1 — xji + (1 — x)4
jln«. (1— aj)f — (1— x)+(l— j;)-| — (1— x)-S+(l— a;)-*— &c
, I , may be expanded by placing
a — bx ss z*, deducing the value of a + bx, and proceeding as beibie.
Take, (p^^ ■ Kace, I — x = s?. Then, x » 1 — 2*, ud
l + x=2 — 2*. Hence, (rz j = -; j-r- — ^ = -5 . . . . .
^ ' M + a:/ 4 — 42:"+ 2* «• — 42*+4
Develop this fraction and replace z by its value (1 — a:)}.
860. It is obvious that there are an infinite number of irrational
expressions that may be developed by first making them rational in
terms of a new variable, expanding the new expression, and replacing
the new variable by its value in terms of the old. So, there are an
infinite number of expressions, involving negative exponents, that may
DERIVATION. 847
be expanded bj substituiiDgi for the old variable, a new variable affected
with » positive exponent, and proceeding oa before. Thus, let it be
ar' x'^
required to expand -= :^-. Let x""* = z. Then, ar^ := z*. Hence,
1 — X"* 1 — 2" '
X-* + X-^ XT* + X-'*» — X-'« +&C.
There are many expressions, however, involving negative and frac-
tioDal exponents, that cannot be developed bj the method of undeter-
mined coefficients.
DERIVATION.
861. When one quantity depends upon another for its value, it is
said to be a function of the quantity upon which it depends. The
dependent quantity is the function, and that upon which it depends is
called the variable. Thus, in the equation, y = 2x, y is the function
and X th^ variable, because, by changing the value of x, y may be made
to have any system of values. Thus, when x = 0, }, 1, 2, 3, &o ;
yss 0,1,2,4, 6, &c., it is evident that, in eveiy single equation involving
two unknown quantities, there is a function and variable. Either of the
unknown quantities may be assumed at pleasure as the function, because
both unknown quantities are mutually dependent upon each other. But
it is usual to regard that unknown quantity as the function, with refer-
ence to which the equation has been solved. If the equation, y = 2x,
is solved with respect to x, we have x = ^. Then, x is the function
and y the variable, and, by assuming arbitrary values for y, x may be
made to have an infinite system of values. The most general form of a
solved equation of the first degree with two unknown quantities is, y =
ax + h. In this, a and b are called constants, because their values
undergo no change. It is plain that y depends for its value upon x, a,
and b, but, as its value only changes with x (since x is the only variable),
y is called a function of x only. If we have a single equation involving
three unknown quantities, any one of these unknown quantities is a
fiinction of the other two.
If we have y =« ax, or y =■ ax + 5, and x = fnz, or x ss mz +ny
848 DEEIVATION.
we call y an explicU function of x, and an implicit fdnction of Zy
because y directly depends upon x for its value, and indirectly upon x.
It is plain that we may have explicit functions of any number of
variables, and, also, implicit functions of any number of variables. We
will, however, confine ourselves to explicit functions of single variables.
If we take the equation, y =s ax + h, and attribute to x arbitraiy
values, 0, 1, 2, 8, 4, &c., and call the corresponding values of y, y, y',
y", y, &c., we will have y = 6, y' = a + 6, y" = 2a + 6, y" = 3a
+ h. Then, y — y = a^f—y*^a, y"' — y" = o, &«. That is,
the difference between any two consecutive states of the function is
constant. This constant increase of the state of the function is called
the differential or derivative of the function. It is evident that, when
the function and variable are both linear, that is, both of the first
degree, the difference between two consecutive states of the function
will always be constant, and then will be truly the derivative of the
function. Thus, take the equation, y = 2a; + 2, and let x have the
constant increment i, beginning at ; that is, let a; =0, + }, i + }»
1 + i> li + } ; ^^6^ ^6 difference between two oonseoutive vahies of
y will always be constant For, we have y=^y ^ = 1 + 2 = S,
y" = 2 + 2 = 4, y"' = 3 + 2 = 5, &c., and, consequently, }f — y =
l=y"-y=y'"-y",&C.
But, whQu the variable is of the second degree, and the function of
the first, the difference between two consecutive states of the funetioB,
corresponding to constant increments of the variable, will not be con-
stant. Thus, take the equation, y=zs^^ and let x have the constaal
increment, 1, beginning at aero. Then, y = 0, y = 1, y" = 4, y = 9,
Ac, and t/ — y = 1, y" — ^ = 3, y"' — y" = 6. The student of
geometry will understand that there ought to be no constant increment
to the function, corresponding to a constant increment to the variable in
the equation, y = x*. For, y expresses the surface of a square of which
X is the side, and the surface of course increases more rapidly than the
side. If the function is of the first degree, and the variable of the
third, it will be seen that the difference between two consecutive states
of the function varies still more widely from a constant. The function
and variable must then both be of the first degree, in order that the
difference between two consecutive states of the function may be eoa-
stant. But, to return to the illustration of the square, it is plain that,
if the constant increment* to the side of the square was ind^nitdy
small, the difference between two consecutive states of the function^
which represents the increment of the function, would be so aearij coo-
DERIVATION. 849
stant that it might be regarded as constant. Suppose yz=s:x?, and that
a; = lioot Then the surface of the square is one square foot. Now,
suppose the side of the square to receive the constant increment,
mm P^^ ^ ^ ^^*- ^^'°' ^ = ^'^ = ^ + 10500 + (100007'
^'^ ^ "^ iOOOO ■*■ (10000)'' *^' ^®°' ^~^ "^ 10000 "^ (lOOOO/
2 2
/' — y = YogoQ + nooom*" ^'^^ ^® ^^ *'^*^ *^^ difference between
the second and first states differs only from the difference between the
third and second, by t^^tj^*
Upon this principle the derivative or increment of a function is
found. It is the difference between two consecutive states of the
fuDction, when these states are indefinitely near to each other, and it
expresses the increment of the function. The word increment is used
in its algebraic sense. When any state of the Unction is less than the
preceding, the derivative is truly a decrement.
To find the derivative of the function, y=z7?, let h represent the
infinitely small constant increment to x. Then, yzz^a?, y'=. (x + A)'
= x' + 2xh -f A*, and y — y = 2xh -J- A". Now, since h is infinitely
small by hypothesis, h* will be an infinitely small quantity of the second
order, and may therefore be neglected. Thus, = — tttt— is veiy small,
but Y^ — zTTi — r-g is much smaller. Now, by making A indefinitdy small,
we have taken the states indefinitely near to each other, and, conse-
quently, y — y is the derivative of y = x*. Hence, the derivative of
cc* is 2xh. It is usual to represent h the increment of the function by
dxy read differential of x. The derivative of the square of any variable
function is then twice the first power of the variable into the differential
of the variable. The differential coefficient of a function is the
difiTerential of the function divided by the increment of the variable,
and is generally expressed by y - In the preceding example the
differential coefficient of x* is 2x ; and, in general, knowing the
differential of the function, we get the differential coefficient by dividing
by the increment or differential of the variable ; and, conversely, we get
the differential of the function from the differential coeffu^ient, by mul-
tiplying by the differential of the variable. It will be seen that^ when
80
360 DERIVATION.
we make ^ =s 0, or indefioitely small, that the differential coefficient
assumes the form of ^, for then ^ = y.
To find the differential coefficient, we hare the following
RULE.
Give to the variable a variable increment^ h, and find the new state
of the function. Take the difference between the new and old Btates of
the function, and reject the terms involving the higher powers of the in-
crement y a% being infinitely smqll quantities of the second, third y d?c.,
orders. Next change h into dx, dy, or dz, (fee, according as the ra-
riable is x, y, or z, dhc. We then have the differential of the function.
Divide the differential of the function by the differential of the va-
riable, and we have the differential coefficient.
Eeqnired the differential coefficient of the function, y s=a^. By the
rule, we have / = (« + A/ = x* + Sx'A + 3A'a; -f h*. HeDce,
y — y = 3a:?A + Bh^x + 7t* ^ 3a;'A, or 3x*c?x, when h is infinitely
small. Therefore, "7^ = ^^^
Newton regarded all algebraic expressions as the representatives of
lines, surfaces, or solids; and supposed lines, whether straight or curved,
to be generated by the flowing of points according to fixed laws, surfiices
to be generated by the flowing of lines, and solids to be generated by
the flowing of surfaces. Thus, a point, moving or flowing according to
the law that it shall always be in the same plane, and at the same dis-
tance from a point, will generate the circumference of a circle ] and a
straight line, flowing with the two conditions of being constantly parallel
to, and at the same distance from a fixed line, will generate the snrfiftce
of a cylinder. A straight line flowing in the same direction with the
condition that, in all its positions it shall continue parallel to itself in
its first position, will generate a plane. One line revolving aixmnd
another, to which it is perpendicular, J_, will generate the surface of a
circle. A flowing square will generate a cube; a flowing semicircle, a
sphere. Newton regarded the increment of the function as expressing
the uniform rate of increase of the function, and called it fluxicn.
The thing generated he called ^uenf. It is evident that the fluxion
does not express the uniform rate of increase, except when the states
are tsken indefinitely near to each other. Suppose a cannon-ball to
leave the mouth of a piece with a velocity of 2000 feet per second, and
that this velocity is reduced to 1200 feet per second at the end of the
DERIVATION. 351
third second. It is plain^ that the difference between the spaces passed
over in any two consecutive instants of time will not be equal to the
difference between the distances passed over in any other two consecu-
tive instantS; unless those instants are inappreciably small. But, for
the millionth part of a second, the velocity might be regarded as con-
stant. If the instant was then taken thus small, the difference between
the spaces in two consecutive instants would be constant.
The differential of the function has, in accordance with the New-
tonian theory, been defined to be the uniform rate of increase or
decrease of the function. The differential of a constant then, must be
zero> since a constant admits of neither increase or decrease.
Theorem I.
362. The differential of the algebraic sum of any number of func-
tions of the same variable is equal to the algebraic sum of their diffe-
rentials.
Let y =^ zt.u±vzt:wzhz; Uy v, w, and z, being functions of the
same variable, x. Give to the variable a variable increment, h, and
call the new states of the function «', v', ti/, and «'. Then y = «' =t:
t/ ± «/ d= 2/, and y — t/ =a tt' — udz(v' — v)± (w' — w) d= (/- — * ^J
as enunciated, since u' — t* is the differential of w, v' — v, the diffe-
rential of V. A shorter notation is dy s= du zh dv ± dw^ dzy and i»
read differential of y, equal to the differential of «, plus or minus tbe
differential of v, &c.
Let y = aa^ -f ex. Then, dy ^ 2axdx -|- cdx.
Theorem II.
363. The differential of the product of a constant by a variable, is
equal to the product of the constant by the differential of the variable.
Let y = Ax. Then, dy = Adx. For, 1/ « A (x + ^)> *^*^
y — y =« AA, or dy = Adx.
Let u sss my, Then^ du = mdy.
Theorem III.
364. The differential coefficient of the power of a quantity i» eq
to the exponent of the power into the quantity affected with an expo-
sent less by unity than the prinaitive exponent.
852 DERIVATION.
Let y = x». Then, ^LlZJ. = nx"-*.
For, give to 2; a variable increment, A ; then, y ss (x -)- A)", and
^—^ = ^ ^ ^ = ^r ,i\ = ' ^y ropreseni-
Q% ^_ ^B
ing X + A by a. But, = a"-* + a^-'x + a'"^x' + a*-^x* +
a — -X
&G., on to » termB. Now, when A is made indefinitely small, x is eqnal
to a, the first member of the equation in y then becomes equal to the
differential coefficient, and the second member becomes a**' + a*~' +
a»~* + &c. s= na^'^'^j as enunciated.
Sy
Hence, we have j- = na*"', and, consequently, dy = naU^^dx, The
differential, as well as the differential coefficient of a power, is then
known.
Theorem IV.
865. The differential of the product of two functions of the same
variable is equal to the sum of the products which arise from multiply-
ing each function by the differential of the other function.
Let y = uv. Then dy = udv + vdu, in which u and v arc both
functions of the same variable, x; the new state of the function is
fl/ = t« + du^ and the new state of the function t/ = i; -f (f i?. Hence,
y = «V = (« + du) {y + dv) = t*» + udv + vdu + dudv. But,
since du and dv are indefinitely small, their product, dudv^ is an in-
definitely small quantity of the second order, and may therefore be
neglected. Hence, xf =.uv + udv -\- vdu, and y — y, or dy =udr
+ vdu, as enunciated.
Corollary.
866. 1st. The differential of the product of any number of functions of
the same variable is equal to the sum of the products which arise from
multiplying the differential of each function by the product of the other
functions.
Let y = 8vmz, in which s, w, u, and z are functions of the same
variable, z. Then, dy ^ uni^xU + 9uzdw + stozdu + svmdz.
To show this, we will first get the differential of the product of three
variables. Let y =z szv. Make 9z=u, Then y = uv, and, from what
has just been shown, dy = udv + vdu = szdv + vd (sz) z=i szdv +
V (tdz + zdi) := szdv + vtdz + vzds, as enunciated.
DERIVATION. 353
Now, knowing the differential of the product of three variable func-
tions, we can readily pass to that of four. For, let y = %wuz. Then,
dy = sumdz + zd (swu) = stcitdz -f- z (sicdv -f- siidw + wuds) = swudz
+ Z8v>du -f Z9udw + ztcuds. And, it is plain that the same process can
be extended to the product of any number of functions.
2d. Since the differential of the product of Av is Adv + vdA =
Adv, A being a constant, we have Theorem II. demonstrated in
another way.
Theorem V.
367. The differential of a fraction b the denominator into the diffe-
rential of the numerator, minus the numerator into the differential of
the denominator, divided by the square of the denominator.
g
Let y = — ,8 and v being both functions of x. Then, dy ss
vds — sdv
g
For, since y = — , we have yv== s; and yv being equal to «, the in-
crement of yv must be equal to the increment of s. That is, d (^yu) =s
- ■» . T T y ds ydv ds sdv vds — sdv
ds, OT ydv + vdy = ds, or dy = ~ = - — — = -g ,
as enunciated. In the expression, - — , we substituted for y its value, — ,
and then reduced the two fractions to a common denominator.
Corollary.
368. 1st. When the denominator is constant, dv is zero, and dy =
3 = — s- = — , equal to the differential of the numerator
TT IT V
divided by the constant denominator. We might arrive at this result
in another way, for, when v =z a constant, — may be written — , and,
1 1«
since — is constant, the differential of — will, by Theorem 11., be
V ' V ' "^
Ids ds
' -, or ~.
V dv
30* X
864 DERIVATION.
369. 2d. When the numerator is constant, ds:=iO, and c?y =
vds — sdv , , sdu . .^
j , becomes ay = j , a negative result.
This ought to be so, for, when the numerator is constant and the
denominator variable, any increment to the variable, Xy upon which v
depends, will decrease y; and, consequently, di/, which expresses the
algebraic increment of y, is tnily a decrement, and ought, therefore, to
have the negative sign. In the case supposed, y is a decreasing fanctioii
of the variable, x, and we see that the differential of a decreasing
function is negative.
We are now prepared to find the differential of any function affected
with any exponent, whether positive or negative, fractional or entire.
Theorem VI.
370. The differential of a quantity affected with any exponent, is
the product arising from multiplying the exponent of the power into
the quantity affected with an exponent algebraically less by unity than
the primitive exponent, into the differential of the variable.
We will first suppose the quantity to be affected with a negatiye
exponent. Let y = i;^ = — . Then, by Corollary 2d, of the last
Theorem, ay = ^^ = ^ = — nv"^ , as enunciated.
r
Next, suppose the exponent to be a positive fraction, y = v*. Then,
f/='^v', or y" = T?'. And, taking the increments of both members,
we have, by Theorem III., since r and s are positive and entire,
_, , , . , , rv^^dv r v^^dv r if^*dv
r v^^dv r v * dv r ^, r ^-i
= 7 X -^ "^ T ^ '"^ = —v*dv = -i;- dv, as
enunciated.
r
Finally, suppose y= t;~»'; we will find by proceeding exactly in the
same way as when the exponent is a positive fraction, dy = ^ x
"* dVf as enunciated.
DERIVATION. 355
Corollary,
370. 1st. The difierential of any parenthetical expression is equal to
the exponent of the parenthesis into the parenthesis^ raised to a power
algebraically less by unity than at first, into the differential of the
quantity within the parenthesis.
This is evident; since we may represent the quantity within the
parenthesis by a single variable, v. I'he parenthetical expression will
then assume the form of v', and may be differential according to the
rule.
Let y se (a. -+• tx*)", place a + fcx' = r, then y = v", and dy =
n^r-^dv ^n(a + ha^y'd (a + 6x*) = « (a + «««)»-' 2hxdx =«
2hn (a + hoFf^^xdx.
372. 2d The differential of a radical expression is equal to the diffe-
rential of the quantity under the radical, divided by the index of the
radical into the radical raised to a power algebraically less by unity than
the primitive index. This is merely a particular case of the preceding,
for a radical is nothing more than a parenthetical expression affected
with a fractional exponent, the numerator of the fraction being unity.
Let v^i7 be the radical; this is equal to v "", and its differential then
^,1 !_,- 1 \=±, dv dv
must be— 1;» dv ss —v ^ dv -=: — 5=r = — , as enunciated,
n n nv-^ nVt;»-»
When n = 2, the radical is of the second degree, and the expression
dv
becomes — =, that is, the differential of a radical of the second decree
2s/v
is equal to the differential of the quantity under the radical sign divided
by twice the radical.
EXAMPLES.
1. Required the derivative of y = i/a -f- a:^.
. - 2xdx
Ans. dy =
3V(a + ^
2. Bequired the derivative of y = v^a + ba?.
. - 2bxdx
An$, dy s
2x/a + 6ac*
8. Bequired the derivative of "syStx + ^^"•
(a + nhx^"^) dx
Ans.
mV(ax + 6x»)--*
356 BINOMIAL FORMULA.
GENERAL EXAMPLES.
1. Required the derivative of a + hx — ca^ + mx*.
Ans. (b — 2cx + Bma^ dx.
2. Required tlie derivative of — -= — .
COCr
cx'hdx ^ (a + hx) 2cxdx
Am.
ex'
3. Required the derivative of rc" (a + hx)^.
Am, m{a '\- 6ic)»x"~' dx -f fiaf (a + &c)"~' hdx.
BINOMIAL FORMULA.
373. By actual multiplication we can readily find
(a + xj == a« + 2ax + oj*,
(a + x)* = a' + 3a^x + Sax" + x",
(a + x/ = a* + 4a*x + 6a«x« + 4aV + x\
We observe a simple law in regard to the exponents in these three
developments. The exponent of a in the first term is the power of
the binomial, and this exponent goes on decreasing by unity unto the
last term, in which it is zero. For, x*, x*, and x* may be written, a*!*,
a^x*, and a^x^. The exponent of x is zero in the first term, and goes
on increasing to the last term, in which it is equal to the power of the
binomial. With regard to the coefficients, we observe that the coeffi-
cients of the extreme terms are unity, that the coefficient of the second
term is the degree of the binomial, and that the coefficient of the third
term can be found by multiplying the coefficient of the second term by
the exponent of a in that term, and dividing by the number of terms
which precede the required term. Thus, in the development of
(a + x/, to find this coefficient, multiply 4, the coefficient of the eecood
term, by 3, the exponent of a in that term, and divide by 2, the number
of terms which precede the required term, the quotient 6 is the co-
efficient of the third term. So, to %id the coefficient of the fourth term
of (a + x)^, multiply 6, the coefficient of the third term, by 2, the ex-
ponent of a in that term; and divide the product; 12, by 3, the number
BINOMIAL 70BMULA. 357
of terms which precede the required term, the quotient^ 4, is the co-
efficient of the fourth term. This remarkable law in regard to the co-
efficients, holds true even for the second and last terms.
We observe; furthermore, that the sum of the exponents of a and x
in every term is equal to the exponents of the binomial and that the
coefficients, at equal distances from the extremes, are equal.
Newton showed that the law in regard to the exponents and
coefficients was general for any binomial of the form (a + x)"", and
111 , . N« - . -_i m(m — l)a"~*a;' .
that we would have (a + x)" == a" + ma'^x H ^^ ^ 1-
m(m — l)(m — 2)a"-^x» .
2T3 "^ "" '
DEMONSTRATION.
374. To demonstrate this usefdl and remarkable Theorem, let us
assume (a + a:)»= A + A'x + A'V + A' V + AToT ; in which
m is taken positive and entire, and A', A", &c., independent of x.
Then, by the principle of undetermined coefficients, we have a right to
make x = in both members of the assumed identical equation.
Making a; = 0, we have A = a"
It was shown that, when two functions of the same variable were equal,
that their differentials were equal. Hence, taking the derivations of
both members of the assumed equation, we have m(a + x)''-^dx ^
(A'+ 2A"x + 3A"V + kc^dx, or, dividing by dx, m(a -f x)-^» =.
A' -f 2A"x 4- 3A'"x» + &c. (N). Now, make a: = 0, and we get,
ma^^^ s A'. Again, taking the derivatives of both members of (N),
and dividing by dx, we get, m(m — 1) (a + x)"*"* « 2 A" +
2.3. A"'x + &c. (P). Making x = 0,we have A'= ^(^ — 1)«""*
Again, taking the derivation of (P), dividing by dx, and making a; ss 0,
-. u km ^('"^ — !)(*"' — 2) a"-* -. ...
-there results A ' = — ^^ iro • ^^ 1^^^ manner, we can
Z . o
find A 2T3T4 •
And A" will plainly be equal to
m(m — 1) (w — 2) . . . . («i— (m — l))a"*'*
1.2.3 (m — 2)(m — l)m ^
ajad, since the factors of the denominator are the same as those of the
nixmerator^ only written in reverse order, we have A*" k 1.
858 BINOMIAL FORMULA.
NoW; replace A, A', A", &o,, by their values ia the afisomed
identical equation; and we have (a + x)" = a" + ma^"* x +
m{m — 1) a»-^ x* m {m — 1) (w — 2) a-"" re'
+ i — o— o r
1.2 • 1.2.3
m(m — 1) (m — 2) . . . (m — n + 2) q»-»+' x'-'
1 . 2 . 3 . 4 . . . (n — 1) "^ *■•
The term,
wi(7n — l)(w — 2) . . . (w — ?i + 2)a«-»+*x»~'
1 . 2 . 3 . . . (n — i) '
ifl called the n^^, or general term. An inspection of the formula will
show that the first factor of the numerator of the coefficient of a in
any term is m, the next factor (m — 1), and so, on decreasing by unity,
unto the last factor, which is m, diminished by two less than the place
of the term. Thus, the last factor of the numerator of the coefficient
of the 4th term is (m — 2). So, the last factor of the numerator of
the coefficient of the n^^ term must be (m — n -f 2). The denominator
of the coefficient of a, in any term, is always the continued product of
the natural numbers, from 1 up to one less than the place of the term.
'Hence, the denominator of this coefficient in the n^^ term must be
1.2.3 (n — 1). In regard to the exponents, we see that a, in any
term, is always affected with the exponent of the binomial, diminished
by the number of the term less one, and the exponent of x, in anj
term, is always one less than the place of the term. Hence, for the n*^
term, we have a"~*+* a;"~*.
The binomial formula is usually written in the ascending powers of a,
instead of x. To develop (x + a)" in the powers of a, it will be
necessary to regard a as the variable, and x as the constant, to make
a = in the successive steps of the operation. Afi we would obviously
obtain the same result, with the exception of an interchange of x and a,
we may at once write
»_i , ni (m — 1) x*"*"" a'
{X + a)" = x" + mx"-* a •\ ^ = — ^ f-
m(m — l)(m — 2)x""*a* m(m — l)...(m — w + 2) x»~"^' a»~'
1.2.3 "*;*• ■ 1 ..2 . 3 . . . (n— 1)
a'
We will repeat the laws in regard to the exponents and coefficients.
1. The first letter of the binomial appears as the first term of the
BINOMIAL FORMULA. 359 i
development; affected witli an exponent equal to that of the binomial,
and the exponents of this letter in the successive terms decrease by
unity unto the last term^ where its exponent is zero. The exponent of
the second letter of the binomial is zero in the first term of the develop-
ment; and one greater in each of the successive terms^ and, in the last
term is m, the exponent of the binomial.
2. The sum of the exponents of x and a, in every term of the
development, is equal to the exponent of the binomial. A simple in-
spection of the formula will show this.
3. The coefficient of the first term of the development is unity; that
of the second term is equal to the exponent of the binomial; that of the
third term is formed by multiplying the exponent of the first letter of
the second term by the coefficient of that letter, and dividing by one
less than the number denoting the place of this required term ; that of
the n** term is formed from the (n — !)*•» term, by multiplying together
the coefficient and exponent of the first letter of that term, and divi-
ding this product by (n — 1), that is, by one less than the number of
the required term.
4. There will always be wi -t- 1 terms in the development. For, we
get one term, a", or x^, without differentiating, and, since the exponent
of m is diminished by unity for each differentiation, it is plain that m
derivatives can be taken of (a -{- sc)", and, since we get a term of the
development by each derivation, we must have in all (m + 1) terms.
6. The coefficients at equal distances from the extremes are equal.
For the developments of (a -|- a;)" and (x -\- a)" must be identical,
differing only in the order of the terms; the first term of the develop-
ment of (a + a;)" being the last term of that of (x -f a)", the second
from the left of the development of (a + a;)" being the second from
the right of that of (x + a)" , &e. Hence, the terms of the develop-
ment of (a -f x)", taken in reverse order, will constitute the direct
development of (x + a)". It is plain, then, that the second term from
the left must have the same coefficient as the second term from tne
right, the one being formed from cc" in the same manner as the other
from a", and so, in like manner, all the other coefficients at equal diB-
tances from the extremes must be equal.
6. The sum of the coefficients of the development of (x + «')'" ]^
equal to the m*>» power of 2. For, if we make x = 1 ^and a = 1- ^^^
»-. . m(m-l) ^""« -I.
the equation, (x -f a)" = x" -f mx"~' a -\ ^^ r^^
860 BINOMIAL FORMULA.
— ^^ ^^ \ ~ ~ .... a", the second member will reduce
to the sum of the coefficients, and the first member will become 2", and we
will have (1 + 1)- « 2- = 1 + m + m(m — 1) + ^(.^—^)0n—2)
+ &c. We make x and a unity, in order to reduce the terms in the
second member to their coefficients; and, it is plain, that if x and a had
another value than unity, the second member would not be the sum
of the coefficients.
In accordance with the demonstration, the sum of the coefficients in
the development (x + a)', ought to be 4 ; for, in this case, m = 2, and
(2)" a= (2)' s= 4. And, in the development of (x + a)', w = 3, and
(2)" = (2)' = 8, which agrees with the fact. So, likewise, in the
development of (x + ay, m = 4, and (2)" = (2)* = 16, which also
agrees with the fact.
FORMATION OP POWERS BY THE RULE.
375. Let it be required to find the development of (x H- of by the
rule. The first term is x*, the second term has 5 for a coefficient, and
the exponent of x is one less in this term than in the preceding -, a
also enters to the zero power in the first term ; and, since the expo-
nents of a go on increasing by unity, a must enter to the first power in
the second term. Hence, the second term is 5x*a. The coefficient of
x in the third term is formed by multiplying 5, the coefficient of a; in
the second term, by 4, its exponent, and dividing the product by 2,
5x4
the number of terms preceding the required term. Hence, -r-^ — as 10,
is the coefficient of the third term, and, from the law of the exponents,
X must enter to the third power, and a to the second power in the
third term. Hence, that term is lOxV. For the coefficient of x in
10 . 3
the fourth term, we have — ^ — = 10, and the fourth term must be
o
10 . 2
lOa^'a'. For the coefficient of the fifth term we have — 3^ — r= 5, and
4
that term itself must be bxa*. For the coefficient of the sixth tenn
5 fl)
we have — ^ = 1, and that term itself must be aPa^, or a*. For the
coefficient of the seventh term we have ' = 0. Hence, ihere is no
BINOMIAL FORMULA. 3G1
seventh term. Then, the development of (a; + a)* is a:* + 5x*a +
lOa^a* + lOxV -f bxa* + a*, and we see that the sum of the coeffi-
cients is equal to (2/ s 32, and that the som of the exponents of x and
a in each term is 5.
The preceding development might have been obtained directly from
the general formula, (x+ajT = ac" + maf^^a H ^^ — a — ^ f- &c.,
by making m as 5. But the above process is generally shorter when
m is positive and entire.
DEVELOPMENT OF (« — o)-.
376. The development of (x — a)" may be obtained in the same
way as that of (x + a)*". But it may be gotten at once from the gene-
ral formula by changing + a into — a ; the terms involving the odd
powers of a will all be negative, and we will have (x — a)" = of* —
"^" + -StT 1.2.3 + =^" •
The last term will be positive when m is an even number, and nega-
tive when m is an odd number.
It will be seen that all the laws in regard to the development of
(x -f a)" hold good in regard to the development of (x — a)", except
that the alternate terms must be affected with the negative sign, and
that the sum of the coefficients is zero.
A few examples will illustrate the use of the two formulas for the
development of (x + a)" and (x — a)".
EXAMPLES.
1. Eequired the sixth power of (x + a).
Ans, af + 6x*a + 15xV -|- 20x»a' -h ISx^a* -|- 6xa« + o^.
For, w = 6, m — 1 = 6, (m — 2) = 4, m — 3 = 3, m — 4 = 2,
m — 5 = 1. Hence, (x + a)" = x" -|- «nx"-* + — ^^ -^
-f .... a" becomes, in this case, (x + oyssx'-fSx* H — '—^ a*
mi
= »• + 6x* -{- 15xV + a^.
2. Eequired the development of (x + of.
Am, x'-f 7x«a -f- 21xV+ 35xV+35x»a< -h21x«a*-H 7xa«-|- a''.
31
862 BINOMIAL FOBMLLA.
3. Develop (x + «)*•
Ans. x^'+Sx^a + 28xV+56a;V-f-70xV+56a:»a'+28x V+8xa'+a'.
4. Develop {x — ay.
Ans. x^ — 6ar^a + ISxV — 20xV + ISx'a* — 6xa« + a*.
5. Develop {x — a)^.
Ans. x' — 7x«a + 21x*a» — 35x*a»+35xV — 21x*a + 7xa« — a\
6. Required the developmeDt of (x — a)'.
Ans. x»— 9x«a + 36xV — 84x«a» + 126x«a* — 126xV + 84x*a*
_ SGx'a^ + 9xa« + a*.
377. The binomial formula has been deduced upon the hypothesis
that the coefficients and exponents of x and a were unity, but it can
be applied to binomials in which these conditions are not fulfiUcd, by
substituting for the letters within the parenthesis other letters vhose
coefficients and exponents are unity. Thus, to get the fifth power of
a* + 3/, let a« = X, and Zf—a; then, {^ + 3y»)* = (x + «)* = Jr'
+ bx*a + lOx'a' + 10x"a' + 6xa* + a', which, by substituting for x and
a their values, will be found equal 2'° + 15ay + QO^y + 270:<y* +
In like manner, we may find the development of (wx» + ra<y^ =
«_i «»_- m (m — 1) 7^n"'-'xP"-*Pa*> _ _ , , ,
^m^prn ^ mm"-*xP»-Pa4 H ^^ ^T 1- . . - r^a**". (A)
In this general formula, make x = 1 , and a = 1 ; the second mem-
ber will reduce to the sum of the coefficients of x, and the first member
will become (n + r)". HencBy the sum of the coefficients of any bino-
mial development is equal to the m'^ power of the sum of the coefficients
within the parenthesis.
In accordance with this general formula, the sum of the coefiicients
of the development of (2' H- 3^*)* ought to be equal to (4)* = 1024;
and this agrees with the development above, for 1 + 15 -f 90 -|- 270
4- 405 + 243 == 1024.
378. Formula (A) shows, moreover, that if the exponents of x and
a are equal, that is, jp = ^, the sum of the exponents of x and a in
each term of the development will pm. Whenever, then, the exponents
ofx and a are equal in any hinomial, the sum, of the exponents of these
letters in each term of the development will he equal to the common ex-
ponent within the parenthesis into the exponent of the power to which
the binomial is to be raised.
BINOMIAL FORMULA. 863
The two important laws just deduced from formula (A) are general
for all binomials. The second and sixth lawS; observed to govern the
development of (x + a)", are but particular cases of the foregoing.
The binomial formula can be extended to polynomials^ by representing
the polynomial by the algebraic sum of two letters. Thus^ to obtain
the square of a;* + 4x" — %x — 8, represent cc* + 4ta? by a, and — 8x
_ 8, by a. Then, (x» + 4x2 — 8x — g^a -= (x + a)* = a:" + 2ax
-f a' = (by replacing x and a by their values), t? + 8x* — 80a;* +
128a; 4- 64. In like manner, to obtain the cube of a;' + 2a; — 4,
let a;" = a;, and 2x — 4 = a; then, (a;* + 2a; — 4)' = (x + af=z o^
4- Sx'a + 3a'a; + a', which, by replacing x and a by their values,
becomes x« + 6a;» — 40x* + 96x — 64.
GENERAL EXAMPLES.
1, Find the coefficient of the twelfth term of the development of
(x + a)", by means of the formula for the n*** term.
in(iii— 1) (m — 2) (fit — 3) (w— 4) (w — 6) (w — 6) (m— 7) (m--8)(w— 9)(m — 10
• 1X2X8X4X6X6X7X8X»X10XU
2. Find the twelfth term of the development of (x + a)*^-
50x49x48x47x46x45x44x43x42x41x40a»x"
Ans,
1x2x3x4x5x6x7x8x9x10x11
3. Find the development of (x + d)^.
Am x» + 20x»o + 190x'V + 1140x"a» + 4845x'V + 15504x«a*
+38760x'V+77520x»V+125970x'V + 167960x'V-|- 184756xV>
+ 167960x«a" 4- 125970.r«a« + 77520x^a" 4- 38760x«a" 4- 15504x*a"
4.4845xV« 4- 1140x'a" 4- 190x«a" 4- 20xa» 4- «""• .
4. Find the development of (a;* 4- 2ax — 4a')*.
Ans, x« 4- 6ax* — 40aV 4- 96a''x — 64a*.
6. Find the 4th power of 3a*c — 2hd,
Ans. 81a*c^ — 2lM^hd + 216aV6*(?' — 96a*cZ^*ci* 4- 165W.
6. Find the cube of x* — 2x 4- 1
An9. x* — 6x* 4- 15x* — 20x* 4- 15x* — 6x 4- 1.
7. Required the coefficient of the sixth term of the development
of (x -f ay,
. 7(7-l)(7-2)(7-3)(7-.4) _ 7.x6.5x4x3 _
1.2.3.4.6 -1.2.8.4.5 ~ ^*
864 BINOMIAL lOBMTJLA.
DEMONSTRATION OF THE BINOMIAL FORMULA FOR ANT
EXPONENT.
379. 1. Let m be supposed negative and entire, then (x + a)"" =r
;, which, by
(x + a)" a;" + maj^-'a + m(m — l)x"~*a* + . . a'
r72
tnally perfonning the division, will be found equal to a;~" — »ix~"*~'a —
m( — m — l)a;--"~'a' m( — m — 1) ( — m — 2)a;~"*~*a*
~ 172 1.2.3 .. . =fc
Pjr*"a", &c. Hence, we have the same law of formation as when the
exponent is positive and entire. We might demonstrate this truth more
rigorously by assuming {x + fl)^" = A + A! a + A"a' + A'"a*, &c.,
and proceeding just as we did when m was positive and entire, r^ard-
ing a as the variable. But it is not necessary to repeat the operadoo.
We have a right to assume the exponents of a to be positive and entire
in all the terms of the development; because, if we expand any frac-
tion whose numerator is a constant, and the first term of w^hose deno-
minator is a constant, all the exponents of the variable will be positive
and entire in the development.
2. When m is positive and fractional. Let m =: ^, and assume
(a + x)^ = A + K'x + k"a? + A'"a^ A-a:-(P). Make a?=0,
and we have A = aa. Taking the derivatives of both members of
(P), and dividing by dx^ we have ^(a -f- a;)« = A' + 2A"x +
PJr-y
3A"V -f &c. (2). Making a; = 0, we get A' = ^7"'. From (2),
there results ^{^ — 1\ (a -f x)!"^ = 2 A" + 2 . 3A'"x + &c. (R).
,?-,
Making a? = 0, we get A" = " (— ~ ^)x~^ ^^^ ^^* ^^^^ '^
suits ^(^ — l) (- — 2) {a + x)T-^ » 2 . 3A'" + &c. From which,
2.8
BINOMIAL FOBMUI.A. 865
Continuing the process and replacing the constants. A, A', &c., hj
their valnes, equation (P) becomes (a '\' x)'i a a<t + ±-a« x +
SI
|(Z_l)af-V+^(|— l)(|-2)af-V ±Pa^-^\-y
TT2 1.2.3
DEVELOPMENT OF BINOMIALS AFFECTED WITH NEGATIVE
AND FRACTIONAL EXPONENTS.
380. When m, the exponent of the parenthesis, (x + «)"? i« posi-
tive and entire, it is evident that successive diflferentiation will even-
tually reduce that exponent to zero, and then the next differential will
be zero. And, since the constants, A, A', &c., of the assumed develop-
ment, have been determined by the successive differentiation of
(x + ay, it is plain that the development will terminate. But, when
the exponent of the parenthesis is negative or fractional, successive
differentiation can never reduce it to zero, and, therefore, an infinite
number of constants can be determined in the equation, (x -f a)"*", or,
p
{x + a) s . . = A -f A'x -f A"x* + A'"x' -|- &c. The series will then
never terminate.
EXAMPLES.
1. -Develop = = (1 — ^)~* ^^^ * series.
X ^"~- X
Ans. l+X'\-o^ + x^ + x* + a^-\'&o.
Malce w = — 1, 1 « a?, and — a = a in the formula (x + a)"" =
- m( — m — l)ar~"^a' «
ar-« — mar^^^a ^^ = — 5 &c.
2. Develop = into a series.
X ^- X
Ans. 1 — x + X* — a^ -{• x* — x*-f &c.
1 -L 7x -I- x*
8. Develop — ^ — — = (1 + 7x + x") (1 -f x)-* into a series.
L "T* X
Ans. 1 -f 6x — 6x* + Sx* — 5x* + 5x* — &c.
4. Develop — : — into a series.
^ a + X
a (J^ a* a^ a*
31*
366 BINOMIAL FOBMULA.
5. Develop v'l -f- a? into a series.
6. Develop s/x + 1 into a series.
7. Develop v/2 =8^/1 + 1 into a series.
Afis, ±(l + i — 4 + T« — Ti5+*«-)-
8. Develop -v/5 ■= \/4 + 1 into a series.
Am, -ii (2 + i — g\ + i\^ — &cO-
9. Develop (w* + n*y into a series.
Ans. ± (m + I m-*n* — ^^m-V + TiB»»"""«" — *«•)•
J. .
10. Develop (a + a) mto a senes.
11. Develop (m» + a') into a series.
■
12. Develop (1 + a^y into a series.
^n«. 1+^ — y+ gj — «•
t)i
13. Develop - into a senes.
m/, x« 3a^ 5a^ 35a^ _ *^ \
Arts, =h -vi-2;^.+8;i4— T6;?"*"i28;;? '^''•/•
14. Develop x/a + x into a senes.
_ I X 2 a^ ^ ^_ii -^+ &c
15. Develop ^l + as into a series.
. , Ix 2x« ex* 21a^4. Ac
^n«. 1 +"5 -25 +125"" 625 "*" *^'
BINOMIAL FOKMDIiA. 867
16. Develop i/1 — x into a series.
^•"•1 + T+ 25 +125+625 + *"•
17. Develop v^2 into a series.
CONSEQUENCES OF THE BINOMIAL FORMULA — SQUARE OF
ANY POLYNOMIAL.
381. We have seen that (a + by = a» -f- 2a6 + 6", that is, the
square of a binomial, is equal to the sum of the squares of its two terms,
plus the double product of those tierms. To find the square of a tri-
nomial ; a + h -{■ c, let a + b = 8, Then, (a + 6 + c)' = (« + c)* =
s» + 2«c + c* = (a + 6)* + 2c(a + i>) + c» = a« + 6' + c« + 2ca +
2cb + 2ab; that is, the square of a trinomial is equal to the sum of the
squares of its three terms, plus the double product of these terms taken
two and two. To find the square of a polynomial of four terms, a +b
+ c-|-cf, leta + &-|-c = / Then, (a- + 6 + c + rf)« = (/ + dy
= t^ -^-2^(1 + cP = a^ +b^ + d" + d^ + 2da + 2db -h 2dc + 2ca +
2cb ■\-2ab; that is, equal to the sum of the squares of all its terms, plus
the double product of these terms taken two and two. Now, it is evi-
dent that the same law will hold good for the square of a polynomial
composed of five terms. For this square is composed of the square of
the first four terms plus the square of the fifth term, plus the double
product of the fifth term by the sum of the first four terms. And,
since the square of the first four terms will give the sum of the squares
of all these terms, plus the double product of these terms, taken two
and two, it is evident that, when the double product of the fifth term
by the sum of the other four, is added to the other double products,
we will have the double product of all the terms, taken two and two ;
and it is plain, moreover, that when the square of the fifth term is
added to the sum of the squares of the other four terms, that we will
have the sum of the squares of all the terms. The law is then true for
five terms, and it is plain that the same reasoning can be extended to
six, seven, and any number of terms. Hence, we conclude, that the
sqttare of any polynomial is equal to the sum of tJie squares of .all tnc
terms, plus the double product of alf the terms, taken two and ivso.
888 BINOMIAL FORMULA.
EXAMPLES.
1. Eequired the square ofa + 6-fc + ^+«-
Ans. a' -f 6" + d» + <? -f e* + 2a5 + 2ac + 2arf + 2ae + 26c +
2bd + 2he + 2cd + 2ce + 2de,
2. Required the square ofa + 6 + c + c£ + c+/-
Am. a» + 6« + c« + cP+e« +/« + 2ah + 2ac + 2ad + 2ae-i-
2a/ + 2hc + 2hd + 2&c + 26/+ 2cd + 2cc+ 2c/+ 2<ic-|- 2df'{-2ef.
CUBE OF ANY POLYNOMIAL.
382. The cube of (a + h), is a* + 3a*6 + 3a6* + 6^
To find the cube of a + 6 -f c, let o + 6 = «. Then, (a + 6 + c^
==(« + c)»==««-f 3«»c + 3«c» + <?=a^ + J» + c» + 3a»6 -f 3a«c +
Zh^c + Sc'a + 3c^5 + 6a&c ; that is, the cube of a trinomial is com-
posed of the sum of the eubes of its three terms, plus the sum of th^
products arising from multiplying three times the square of each term
into the first power of the other terms, plus six times the product of
all the terms, taken three and three.
By pursuing precisely the same course of reasoning that we Iwt«
already employed, it can easily be shown that the law is general for any
polynomial.
EXAMPLES.
1. Required the cube of a + 6 -t- c + c?.
Ans, a* + 2»' + c" + dl» -f 3a^6 + 3a"c + Za^d + 3&^a + ZVc +
3&«c^ + 3c«a + 3c^6 + 3c«rf + 3cf»a + 3<?6 + 3d«c + ^ahc + %ahd +
^acd + %hcd.
BINOMIAL FORMULA. o69
EXTRACTION OF THE Nth ROOT OF WHOLE NUMBERS AND
POLYNOMIALS.
383. The most important consequence of the binomial fonnnla is,
that we are enabled by it to extract high roots of whole numbers and
polynomials. We will begin with the simplest case, the extraction of
the n*' root of whole numbers.
EXTRACTION OF THE Nth ROOT OF WHOLE NUMBERS.
Let a represent the tens, and h the units of the required root. Then,
the given number will be (a •\- ft)", and from the binomial formula we
have
(a+6)»=a« + «a-'ft + -^^j— -^- + -^^ TrTTl +*^'
That is, a number, whose n*"* root is composed of tens and units, is
made up of the n** power of the tens, plus n times the (n — l)"* power
of the tens into the first power of the units, plus, &c.
It is evident that the n** power of the tens will give, at least, « + 1
figures, therefore, the n*^ root of the tens cannot be sought in the n
right hand figures. These must, therefore, be cut off from the right.
We next seek the greatest n^^ root contained in the left hand period,
and, when we have found it, we will have a of the formula. Raise the
root found to the n^^ power, and subtract this power from the left hand
period. The remainder will correspond to na'^~^b + — — ~^ -f
&G., of the formula. The approximate divisor to find h (the unit of
the root) is nd^^. The true divisor is na^^ -| ^^ — = — h
— ^^ J-^ — o~^ V &c ; but, as 6 is unknown, we can only use
1 . 2 . o
the approximate divisor. The n — 1 figures, on the right of the remainder,
must be separated from the other figures, because n times the (n — l)***
power of the tens will give, at least, n figures. Dividing the period on
the left by the approximate divisor, we get the units of the root, or, gene-
rally, a number greater than the units, because our divisor is too small.
It is plain that the number of terms rejected when using the approximate
divisor, depends upon the number of units in n, and that the value of
each term depends mainly upon a. When, therefore, n and a are both
large, or when one only is large, the approximate divisor is very consider-
ably too small, and the quotient, therefore, will be too great. Raise the
Y
370 BINOMIAL FORMULA.
two figures qf the root found to the n^^ power, and compare the result
with the given numher ; if it be greater than this number, the last figure
of the root must be reduced bj one, or more. Let us illustrate by an
example.
Required the fifth root of 33554432.
385 54432
243
na^-' = 5 X 81 = 405 92 54432
32
We began by cutting off the five right hand figures. The next step
was to find 3, the greatest fifth root contained in 335, the left hand
period. Next, the approximate divisor, 405, was formed, and this was
found to enter twice in the left hand period of the remainder, after
subtracting the n** power of the tens from the given number. Upon
trial, 32 is found to be the true root, for, (32)* = 33554432. In this
case, the unit figure of the root not being large in comparison with the
tens, the approximate did not differ so materially from the true as to
give a quotient figure too great by unity, or some other number. But,
suppose it were required to extract the fourth root of 230625.
Then, 39*0625
16
25
na"^' =4 X (2)» = 32, 230 625
The approximate divisor gives 7 as a quotient ; but, upon trial^ it
has to be diminished by 2 ; and 25, not 27, is the true root.
It will be seen that the units of the root can only be ascertained by
trial. If the units of the root be large in comparison with the tens,
the quotient obtained by dividing the left hand period of the remainder
by the approximate divisor, will often differ considerably from the true
units of the root. The following example will illustrate this :
</
* ' 13 0321 = 19
wa--' = 4 . P = 4 120321 = na^'b + ^ (^ l)a^b* ^ ^
JL . Z
In this example, the approximate divisor gives 30 as a quotient, but
this is absurd. We try 9, the greatest figure of the units, and find it
to be right; for (19)*= 130321. The quotient, 30, being so large, we
concluded that the units of the root must be large, and, therefore,
tried 9.
BINOMIAL rORMULA. S7l
Onr reasoning has been confined to numbers wbose root contained
but two figures ; but, as in the corresponding cases of the extraction
of the square root and cube root of whole numbers, it can readily be
extended to numbers whose root contains any number of figures. It is
not necessaiy to repeat the generalization of the principles, and we,
therefore, pass at once to the rule for the extraction of the n^ root of
whole numbers.
RULE.
I. Divide the given number into periods of n figures each, begin-'
ning on the right. Extract the n* root of the greater n"* power con-
tained in the left hand period, and set this root on the right, after the
manner of a quotient in division. Subtract the n^ power of the root
90 found from the left hand period,
n. To this remainder anne^ the next period, and separate from the
new number so formed then — 1 figures on the right. Regard the lefi
hand period as a dividend.
m. Divide the dividend by n times the (p. -^y^ power of the first
figure of the root. The quotient will be the second figure of the root,
or a number greater than the true second figure of the root. Raise the
iwo figures of the root found to the v^ power. If the result exceed the
two left haTid periods of the given number, the last figure of the root
muxt be reduced until the n^ power of the two figures is equal to, or
something less than the two left hand periods.
IV. Ann^t the third period to the remainder, after subtracting the
13?^ power of the first two figures of the root from the two left hand
periods, and cut off n — 1 figures from the right of the new number
thus formed, and regard the left hand period as a new dividend.
V. Divide this dividend by n tirnes the (n — Vjf^ power of the first
tiDO figures of the root, the quotient wiU be the third figure of the root,
or a number greater than the third figure. Ascertain, by trial, whether
the last figure is correct, and proceed in this way until all the periods
are brought down. The number of figures in the root will always be
equcd to the number of periods in the given number.
EXAMPLES.
1. Kequired the y 68574961. • Ans. 91.
2. Required the ^6240321461. ^ns. 91,
372 BINOMIAIi rOBMVLA.
8. Required the (/2476099. Am. 19.
4. Required the V 108248216. Am. 102.
5. Required the {/ 1092662195786551. An$. 1111.
6. Required the t/5392 18609632. An$. 222.
7. Required the ^35831808. Ans. 12.
8. Required the ^587008342272. Am. 48.
9. Required the ^75154747810816. Ans. 96.
10. Required the V54165190265169632. Am. 2222.
APPROXIMATE ROOT OP AN IRRATIONAL NUMBER TO
WITHIN A CERTAIN VULGAR FRACTION.
884. The principles of approximation have heen so fully explained
under the head of the Square Root and Cube Root^ that it will only be
necessary now to give the rule for the approximate n^ root to within a
vulgar fraction, whose numerator is unity.
RULE.
Multiply and divide the given number by the n*^ power of the deno-
minator of the fraction that marks the degree of approximation. The
root of the numerator of the new fraction thus formed^ to within the
nearest unit, divided by the exact root of the denominator^ will he the
approximate root required,
EXAMPLES.
1. Required (/6 to within }. Ans, |.
For, vr= v6orr| = vM=i
2. Required ^9 to within }. Am. |.
3. Required 1/^ to within }. Am, f
4. Required 1/^98152 to within f Am. <f, or 8, neariy.
5. Required ^^8589929092 to within f Am. V; or 8, nearly.
BINOMIAL VOBMULA. 873
APPBOXIMATB ROOT OF WHOLE NUMBERS TO WITHIN A
CERTAIN DECIMAL.
RULE.
385. Annex as many periods of ciphers (each period oonnsting of "A
^figureSf) as there are places required in the root. Extract the n*"* root
of the new number ihtis formed to within the nearest unit, and point
from the right for decimals, the numJ)er of places required in the root.
EXAMPLSS.
1. Reqiiired i/Wto within .1. Ans. 2.1
2. Required ^116857201 to within .1. Ans. 41.1.
3. Required ^Z 9998989 to within .01. Ans. 9.99.
4. Required t/268991 to within .1. Ans. 12.1.
5. Required t^ 611. 999999999999999888 to within .01.
Ans. 1.99.
APPROXIMATE Mh ROOT OF A MIXED NUMBER TO WITHIN
A CERTAIN DECIMAL.
RULE.
386. Annex ciphers, if necessary, until the decimal part vnll *^^^^^
a« many periods of n figures each as there are places required *^ ' >
BxtraU the n*^ root, and point from the right the required rvart^ er
decimal places.
EXAMPLES.
1. Required ^TMSi to within .1. Ans. l-^' x^eai?-
2. Required i/2-4884"to within .1. ^»«-
3. Required V3.583181 to within .1.
4. Required i/62403.21461 to within .1.
6. Required ^392.6430946993 to within .01.
j^^rt,^-
6. Required <t/1725.4996508234 to within .01.
32
I. 9.1-
3.31.
4.44.
874 PERMUTATIONS AND COMBINATIONS.
APPROXIMATE Nth ROOT OF NUMBERS ENTIRELY DECIMAL
RULE.
887. Annex ciphers^ if necessary, until there are as many periods of
D Jigures eaxh as there are places required in (he root. Extract the n*
root of the new number iJmsformedy and point off the required number
of decimal places.
EXAMPLES.
1. Required i/.00246 to within .1. Ans. .3.
2. Required V 0028629161 to within .01. Ans. .31.
8. Required V -0004084201 to within .01. Ans. .21.
4. Required ?/.61676701935 to within .01. Ans, .91.
5. Required ^Z. 96059690 to within .01. Ans. .99.
6. Required ^1245732577 to within .01. Ans, .66.
7. Required </ -464404086785 to within .01. Ans, .88.
8. Required li/.0000000000285311670612 to within .01.
Ans. .11.
9. Required V/.00048828126 to within .1. Ans. .5.
10. Required Ii/.000000000000000000000285311670612 to with-
in .001. Ans. .011.
11. Required t/.2706784158 to within .01. ^fw. .77.
PERMUTATIONS AND COMBINATIONS.
888. Permutations are the results obtained by writing n letters m
sets of 1; 2; 3^ .... or n letters, so that each set shall differ from all
the other sets in the order in which the letters are taken. Thus, the
permutations of the three letters, a, h, and c, taken singly, are a, 6, c;
and in sets of two letters each, ah, ac, ha, be, ca, cb ; and in sets of
three letters each, abc, a^b, bac, bca, cba, cab.
PERMUTATIONS AND COMBINATIONS. 375
It will be seen that the sets differ only in the manner in which the
letters are written^ the letters in all the sets being the same.
389. Let it he required to determine the number of permutations of
m letters^ taken n in a set.
Suppose the letters to be a, h, Cy d, &c., and let ns first permute them
singly, we will evidently have m permutations of the m letters taken
one in a set. Now, let us reserve a, there will remain m — 1, letters
b, c, df &c., which, permuted singly, will give (m — 1) permutations, 6,
O ^y ^) fi &<5. : write a before each of these letters, and we will have
m — 1 permutations, aby acy ad, &c., of two letters, with a as the first
letter of each set. Next, let us reserve b out of the letters, a, 6, c, d,
&c., and then permute a, c, <^, &c., singly. We will again have (m — 1)
permutations of the letters, taken in sets of one letter each. Writing b
before each of these sets, we will have (m — 1) permutations of m
letters, taken in sets of two, with h as the first letter. Reserving c in
like manner, we can again form (m — 1) permutations of m letters,
taken two in a set, with c as the first letter of each set. Beserving all
the m letters in succession, we can evidently form m(m — 1) permu-
tations of m letters, taken 2 in a set. And, of these, a will be the first
letter of (m — 1) sets, b the first letter of (m — 1) sets, c the first
letter of (m — 1) sets, and so on. It is plain, moreover, that a will
not only be the first letter of (m — 1) sets, but that it will also be the
last letter of (m — 1) sets, and that it therefore has been made to
occupy all possible positions. As the same remark may be made of b
and all the other letters, it is obvious that m(m — 1) truly expresses
the number of permutations that can be made of m letters, taken two
in a set. That is, the number of permutations of m letters, taken two
in a set, is equal to the total number of letters into the total number of
letters, less one.
In like manner, we may find the number of permutations of m letters,
taken three in a set. For, if we omit one of the letters, as a, there
will be m — 1 letters left, and these, permuted in sets of two letters,
will give (m — 1) (m — 2) permutations. For, we have just seen that
the number of permutations, taken in sets of two, was expressed by the
number of letters into the number of letters, less one. Writing the
reserved letter, a, before each of these (m — 1) (m — 2) sets, we will
form (m — 1) (m — 2) permutations of m letters, taken three in a set,
with a as the first letter of each set. Beserving in succession each of
876 PERMUTATIONS AND OOHIBINATIONB.
the m lettersy we can plainly form m(m — 1) (m — 2) pennntations
of m letters, taken three in a set.
By the same course of reasoning it can he readily shown that the
numher of pei mutations of m letters, taken four in a set, will be ex-
pressed by m (m — 1) (m — 2) (m — 3) ; and of m letters, taken five
in a set, by m (m — 1) (m — 2) (m — 3) (m — 4).
We observe, in all these expressions, that the first factor is the number
of letters, and that the last factor is the number of letters, diminished
by the number in a set less one. Moreover, each factor after the first
is one less than the preceding factor. Ilcnce, the number of permu-
tations of m letters, taken n together, will be expressed by m (m — 1)
(m — 2) (m — 3) (ni — 7t -f 1), or, denoting by A the num-
ber of permutations of m letters, taken n in a set, we have Ass
VI (m — 1) (w* — 2) . . . , , (m — n + 1).
To apply this formula, it is best to determine in the first place the
last factor, we then know where to stop. Since the first fiictor is
always the number of letters, and the law respecting the mean factors
is known, the formula can then be readily applied.
EXAMPLES
1. Required the number of permutations of 10 letters, taken 7 and 7-
Ans. 10(10 — 1) (10 — 2)(10 — 3)(10 — 4)(10 — 5)(10— 6),
orl0x9x8x7x6x5x4 = 604800.
For, n = 7, and m = 10, then m — n + 1 = 10 — 7 + 1 = 10 —
6, the last factor is then 4, and the first 10, the intermediate factonB can
easily be formed.
2. Kequired the number of permutations of 5 letters, taken 4 and 4.
Am, 5(5 — 1)(5 — 2)(5 — 3) = 5x 4 X 3x2 = 120.
3. Required the number of permutations of 12 letters, taken 10 and 10.
Ans, 12 (12 — 1) (12 — 2) (12 — 3) (12 — 4) (12 — 5) (12 — 6)
(12 — 7) (12 — 8) (12 — 9).
4. Required the number of permutations of 12 letters, taken 11
and 11. Ans. 12x11x10x9x8x7x6x5x4x3x2.
5. Required the number of permutations of 12 letters, taken 12 in
a set.
Ans, 12 (12 — 1) (12 — 2) (12 — 3) (12 — 4) (12 — 5) (12—6)
(12 — 7) (12 — 8) (12 — 9) (12 _ 10) (12 — 11) ; or, reversing the
factors, 1x2x3x4x5x6x7x8x9x10x11x12.
PXEMUTATIONS AND COMBINATIONS. 377
Bemarks,
890. If, in the general formula, A=:m(m — 1) (m — 2) . . . .
(m — n + 1); V6 make m = n, and call B the corresponding value of
A, we will have B = n (n — 1) (n — 2) .... 4 x 3 x 2 X 1 ; or,
reversing the order of the factors, B=slx2x3x4 (n — 2)
(n — l)n. Hence, the number of permutations of n letters, taken all
together, b equal to the product of all the natural numbers from 1 to
n, inclusive.
Example 5 is an illustration of this.
It is plain that, by increasing n (m remaining constant), we will in-
crease the number of permutations, because we will have more factors
in the product that expresses the number of permutations. And, when
n is made equal to m, this product is the greatest possible. Thus, the
result is greater in example 4 than in 3, and greater in 5 than in 4.
When n == m + 1, the whole formula reduces to zero. This plainly
ought to be so, since it is impossible to permute m letters in sets of
9n + 1 letters. Zero, here, as in many other places, is the symbol of
impossibility.
When n = I, the last factor, m — n -f 1, is equal to m. Hence,
m
the first and last factors are the same, and we have A = rr- ss m ; that
is, the number of permutations of m letters, taken one in a set, is equal
to m,
COMBINATIONS.
391. Combinations are the results obtained by writing any number
of letters, as m, in sets of 1 and 1, 2 and 2, 3 and 3, .... n and n,
tn and m, so that each set shall differ from all the other sets by at least
one letter. In permutations, the sets differ in the order in which the
letters are written ; in combinations, the sets differ in the letters them-
selves. Thus, the letters, abc, taken all together, give but one combi-
nation, abc ; but will give six permutations,
abc, acb, bac, bca, cab, cba.
If the same letters are taken two and two, they will give but three
combinations
db, ac, be,
wliile each combination is susceptible of two permutations ; and there
32*
378
PERMUTATIONS AND COMBINATIONS.
are, therefore^ six permutations of the three letters, taken two and two,
thuS;
}aby . ) aCf , ) hcj
y ac will give > and be will give r ,
892. Let it be required to determine the total number of combina-
nations of m letters, taken n in a set.
, It is evident, from what has been shown, that if we knew the total
number of combinations of m letters, taken rt in a set, that we ooold
get the number of permutations of m letters taken n in a set, by mnl-
tipljing the number of combinations by the number of permutations
obtained by permuting each combination ; or, in other words, by multi-
plying the number of combinations of m letters, taken n in a set, by the
number of penjiutations of n letters, taken all together. Conversely,
when the number of permutations of m letters, taken n in a set, and
the number of permutations of n letters, taken all together, are known,
we can, by dividing the former by the latter, determine the number of
combinations of m letters, taken n in a set.
893. To iUustrate more fully by an example.
Let us combine the four letters, a, 5, c and d, in sets of three, we
will have the four combinations,
abcy abd, acd, bed.
If, now, we permute each of these sets, taking all the lett^^rs in each
set, we will have the total number of permutations of four letters, taken
three in a set, that is, the total number of permutations of m letten^
taken n in a set.
Writing the results in tabular form, we have
Combinations of a. ft. c. d, in sets of 3
ahc
ahc
acb
hac
bca
cab
cba
abd
abd
bad
bffa
dab
dba
adb
acd
hi^
Permutations of each combination, taken 3
and 3, or all together. |
acd
adc
cad
cda
dac
dca
hcti-
bile'
cbd'
cdh^
dbc'
deb
We see that each of the four combinations, in sets of 3, gives six
permutations, taken 3 and 3. There will, therefore, be 4 X 6 = 24
permutations of four letters, taken three in a set. Now, as a corre^
ponding table could be formed for any number of letters, it is plain that
EXAMPLES.
1. Eequired the number of combinations of 6 letters, taken 4 in a
. 6(6-1) (6-2) (6 — 3) 6.x5x^ jc3_^5^
TTYTsTi — 1.2.3.4
For m = 6 and « = 4. Hence, the first factor of the numerator
6, and the last m — n-h 1 = 6 — 4 + 1 = 6 — 8. The two mo
factors of the nnmerator are then readily formed, the extremes bexi
icnown.
PEBMUTATIONS AND COMBINATIONS. 379
the total number of permutations of m letters, taken n in a set, is equal
to the number of combinations of m letters, taken n in a set, multi- I
plied by the number of permutations of n letters, taken all together. [
Hence, if
X = the number of combinations of m letters, taken n in a set ;
Y SB the number of permutations of n letters, taken all together;
Z = the total number of permutations of m letters, taken n in a set,
we shall have Z = X . Y. Hence, X = ^.
But Z and Y are already known.
For Z = m(m — 1) (m — 2) (m — n + 1), Formula(A)j
and Y = 1 . 2 . 3 . 4 . 6 n, Formula (B). Hence, we have X=
m(m— ) . . . . (m — n -f ) ^^^^ which, we conclude that the
number of combinations of m letters, taken n in a set, is equal to the
number of permutations of m letters, taken n in a set, divided by the
number of permutations of n letters, taken n in a set.
In the application of this formula, it will be of service to remember
that the first factor of the numerator is the number of letters, that eaou
successive factor is one less than the preceding, and that the last fiictor
is the number of letters diminished by the number in a set, less on®-
It is well, then, to determine first the extreme factors of the numerator 5
the mean factors can then be readily supplied.
In regard to the denominator, the last factor ia the number in a se 9
and the factors counted to the left go on diminishing by unity uix*^
the first factor, which is always unity.
380 PEBMUTATIONS AND OOMBINATIONB.
2. Required the number of combinations of 12 letters, taken 7 and 7.
12(12 — 1) (12 — 2) (12—3) (12—4) (12 — 5) (12 — 6)
^'"' 1.2.3.4.6.6.7
12x11x10x9x8x7x 6 ^ 12 X 11 X 10 X 9 X 8
"" 1x2x3x4x5x6x7 ~ 1x2x3x4x5
X 9 X 8 = 792.
= 11
8. Required the number of combiDations of 12 letters, taken 5 and 5.
12 (12—1) (12—2) (f2— 3) (12—4) _ 12x11x10x9x8
'• 1x2x3x4x5 "~ 12x10
= 792.
4. Required the number of combinations of 12 letters, taken 9 and 9.
12(12—1) (12—2) (12—8) (12--4) (12—6) (12—6) (12—7) (12—8)
'• 1.2X3X4X6X6X7X8X9 ""
12x11 X10X9X8X7X 6x5x4 __ 1320 _
1X2X8X4X6X6X7X8X9 "" 6 " '
5. Required the number of combinations of 12 letters, taken 3 and 3
^^ 12 (12-1) (12-2) ^ 12x11x10 ^3^^
1 . J . o o
6. Required the number of combinations of 10 letters, taken 9 and 9.
Ans. 10.
7. Required the number of combinations of 10 letters, taken 1 and 1.
Aiu. 10.
8. Required the number of combinations of 25 letters, taken 21
and 21. Ans. 12650.
9. Required the number of combinations of 25 letters, taken 4 and 4.
An9. 12650.
10. Required the number of combinations of 50 letters, taken 47
and 47. Ans. 19600.
11. Required the number of combinations of 50 letters, taken 3 and 3.
Ans, 19600.
Scholiiim.
394. The last ten examples show that the same number of lettersy
combined differently, may produce the same number of combinations,
that is, the number of combinations of m letters, taken p in a set, may
be equal to the number of combinations of m letters, taken ^ in a set.
We propose to show that this will always be so when m &= j9 + ^. ThuSy
PSEMUTATIONS AND COMBINATIONS. S81
in example 10, m » 50 and p s 47; in example 11, m a 50 and
9 "= 3. The results Lave been the same in those examples, because
m rsp ^ q.
To show this, suppose p'^q- Then the number of combinations
of m letters, taken p in a set, will be expressed thus :
jv __ wt (m — 1) .... (m — q-i- 1 ) (m — q) (m — q — 1) (m—p+l)
1.2.3 (p — 1) JP
And for the number of combinations of m letters, taken qiatL set,
we will have the formula,
r ^(^ — 1) (w* — 2) (m — 9+1)
™ 1.2. S~(q^^~q •
Hence, D = C (m-g) (m-q-1) (m^p + I)
(5^ + 1) (? + 2) p
Now, it is plain that D will be equal to C, when (m — q) (m — q — 1)
&c. =^p{p — 1) . . . . (j + 2) (^ + 1). Or, since the fiictors in
both members go on decreasing by unity, and since the number of
tbem is also equal, the last equation will be true when m — q ^=py or
fn ^» q •{• p, BB enunciated.
This rule is of importance whenever jj> — .
For, then, we have only to take the difference between p and m, and
the number of combinations of m letters, taken m — p and m — p, will
be the same as the number of combinations of m letters, taken j9 and^.
Thus, in example 10, instead of taking the number of combinations of
50 letters, taken 47 in a set, we may take the number of combinations
of 50 letters, taken 50 — 47, or 3 in a set.
In like manner, the number of combinations of 100 letters, taken 90
in a set, would be the same as the number of combinations of 100
letters, taken 10 in a set.
395. If, in the formula, D = C (^)(^"^-l);-(^^-fl)
we make q = p — 1, we will have D = C ^ — ; for, in that
case, D will have only one more fector than C. Hence, ike number of
combinations of m letterSy taken p and p, is equal to the number of
combinations of m letters, taken p — 1 and p — 1, mvltipUed by the
factor When p =3 2, 3, 4, &c., or the sets are made up
of 2, 3, 4, &c., letters, the factor — will become — - — ,
882 PERMUTATIONS AND COMBINATIONS.
. Suppose, for example, that we have 8 letters, or
3 ' 4
in = 8, then the successive multipliers will be J, |, |, J, |, |. AbcI,
since the number of combinations of 8 letters, taken 1 and 1, is 8, we
will have the number of combinations, taken 2 and 2, 3 and 3, 4 and 4,
Ac, expressed 8 X ^ = 28, 28 X | = 56, 56 X | = 70, 70 x t =56,
56 X § == 28, 28 X ^ = 8.
This remarkable law has had many important applications, one of the
principal of which is, the determining of the coefficients in the binomial
formula.
It is plain from what has been shown that, if we write in succession
the number of combinations of m letters, taken 1 and 1, 2 and 2, . . . .
n and 9t, we will have a scries of numbers increasing to the middle term,
and then repeated in retrograde order.
Thus, 6 letters combined, 1 and 1, 2 and 2, 3 and 3, 4 and 4, 5 and 5,
give the series 6, 15, 20, 15, 6. In like manner, 7 letters give the
series, 7, 21, 35, 35, 21, 7. The law of formation is precisely that
which we have observed in the binomial development; and, in £fict, in
this development, the coefficients, after the second, are the combinatioDf
of m letters, taken 2 and 2, 3 and 3, &c., . . . . n and 72.
396. Let us now seek the greatest term of the series formed by com-
bining m letters, 1 and 1, 2 and 2, . . . . fi and n. The factors, which
. ^ . ^, . . m — 1 m — 2 m — n -f 1
determine the successive terms, arc — ^ — , — ^ — , .... ,
Z 6 n
&e. Now, since these numerators go on decreasing, and the denomi-
nators increasing, it is evident that the successive products will go od
increasing until m — n + 1 s=sn, and after that will decrease and be
repeated in reverse order. Suppose m an odd number, then placing
m — n -f- 1 ^ w, we get n = i (m 4- 1). Now, if we include unity,
the last term of the series, there will be m terms in all, and in case of m
being odd, the n^^ term will obviously be even, and, of course, have an
even number of terms preceding and succeeding it. Thus, in the seriess
7, 21, 35, 35, 21, 7, 1 ; n = — - — = 4, will represent the 4th term,
and we see that it has three terms before and three after it. It is pUdO|
moreover, that the n** term, being formed from the (n — 1)<* term bj
multiplying the latter by = !> is? alsO; equal to the (n — 1 )«»
term.
Hence, when m is an odd number, there will be two equal eentnl
PEBMUTATIONS AND COMBINATIONS. 383
tcrmS; the series will go on increasing unto the first of these, and
decreasing from the second unto the last term, unity. Thus, the com-
binations of 5 letters, taken 1 and 1, 2 and 2, 3 and 3, 4 and 4, 5 and
6, give the series 5, 10, 10, 5, 1.
397. When m is even, the factors increase until n=:^im. We can-
not, in this case, have n = — - — , for that would make n fractional, a
manifest absurdity. No factor, in this case, is unity, and, of course,
there are no equal central terms. The terms go on increasing until
n = im, and are then reproduced in reverse order. It is obvious that
whenever w ^ i m, the factor will become a proper fraction,
n
and, of course, the term of the series formed by multiplying the pre-
ceding term by this factor, will be less than the precediDg. Ten letters,
taken 1 and 1, 2 and 2, &c., give the series, 10, 45, 120, 210, 252,
210, 120, 45, 10, 1.
398. If we add together the values of C and D, as found in Art.
394, and at the same time make j = p — 1, we will have + D =3
C + 0(^— j> + l) ^ G(m + l) ^ m + 1 ^^ ^l^ / ^ + 1 \
p p 1 p \ 1 /
(m\ /m — 1 V (m — p + 3) (m — p -f 2)
T/ \ 3 / p — l p '
The second member of this equation plainly denotes the number of
combinations of m + 1 letters, taken p and p, whilst the first member
denotes the sum of the combinations of m letters, taken p and p, and
p — 1 and p — 1. This important relation enables us to get the number
of combinations of m -|- 1 letters from that of m letters. Thus, make
m = 4, p = 3, and p — 1 = 2, then 4 letters, taken 2 and 2, give 6
combinations, and taken 3 and 3, give 4, and we find that 5, or m -|- 1
letters, taken j7, or 3 in a set, give a number of combinations equal to
4 + 6, or 10.
399. Upon this principle, the following table (p. 384) has been con-
structed. The first vertical column is made up of ones ; the second column
contains the natural numbers from 1 to 20, and expresses the number of
combinations of these numbers, taken 1 and 1; the other vertical
columns express the combinations of the same numbers, taken 3 and 8,
4 and 4, 5 and 5, &c. These vertical columns result from the hori-
zontal, which are constructed according to the principle demonstrated
in Article 398, observing that every horizontal column must close with
unity. Thus, the numbers in the second horizontal column are 1, 2
384
PERMUTATIONS AND COMBINATIONS.
and I, the numbers in the next eolamn are, 1 + 2=3, 2+ 1=3
and 1, and express the number of combinations of three letters, taken
1 and 1, 2 and 2, and 3 and 3. Now, prefix unity, and we have the
third horizontal row made up of 1, 3, 3 and 1, and the fourth row will
be made up of 1, 3 + 1 = 4, 3 + 3 = 6, 3 + 1 = 4 and 1, the last
four nlimbers expressing the number of combinations of four letters,
taken 1 and 1, 2 and 2, 3 and 3, and 4 and 4. All the other horizontal
columns are formed in the same manner.
1
1
ARITHMETICAL TRIANGLE OF PASCAL.
1
2
3
4
5
1
1
3
6
10
1
4
10
1
5
1
6
7
8
9
10
15
21
28
36
45
20
35
56
84
120
15
35
70
126
210
6
21
56
126
252
1
7
28
84
210
1
8
36
120
1
9
45
1
10
1
11
55
165
330
462
462
330
165
65
11
12
66
220
495
792
924
792
495
220
66
13
78
286
715
1287
1716
1716
1287
715
2S6
14
91
364
1001
2002
3003
3432
3003
2002
1001
—
15
105
455
1365
3003
5005
6435
6435
5005
300.3
16
120
560
1820
4368
8008
11440
12870
11440
$008 .
1
17
136
680
2380
6188
12376
19448
24^10
24310
19448 1
18
153
816
3060
8668
18564
31824
43758
48620
4S7SS
19
171
969
3876
11628
27132
50388
75582
92378
92378
20
190
1140
4845
15504
38760
77520
125970
167960
184756
landl
2 and 2
3 and 3
4 and 4
5 and 5
6 and 6
7 and 7
8 and 8
9 and 9
10 and 19
400. We can now readily find the entire sum of the combinations
formed by any number of letters, taken in every possible way. Let the
numbers in the t^^ row be expressed by 1, a 6 ... m, m ... 6, a, 1 ; t
being supposed an odd number. Then their sum will be 2 (cr -f 6
-f m + 1). The numbers in the next row will be expressed by 1 + a,
a + ft....+.4-m, 4-m + wi...ft4"07a + l and 1, and their 8am
will be 4 (a + h . , . + m + Y) +1. Hence, the sum of the numbers in
the (^ + 1)*^ column, will be double of the preceding, and one more.
When < is an even number, the same law can be shown to be true
But the sum of the numbers in the second horizontal column, is 2* — 1
Hence, in the third column, it will be 2 (2* — 1) + 1 «= 2* — 1, and
in the fourth, 2 (2'— 1) + 1 = 2* — 1. And it is plain that, in the
^ column it will be 2* — 1. It will be seen that the ones in the fiiBl
PBBMUTATIONB AND COMBINATIONS. 3^5
vertical column are omitted. Hence, 2^ — 1, expresses the sum of the
combinations of t letters, taken 1 and 1, 2 and 2 .... t and t. Thus,
to apply the formula, let it be required to determine the number of
combinations of three letters, taken 1 and 1, 2 and 2, 3 and 3. Then
t =3, and 2* — 1 = 2» — 1 « 8 — 1 = 7. This agrees with the
fact, for the three letters, a, b, c, give us the seven combinations, a,
by c; ah, acy be, and abr. So, likewise, the four letters, a, 6, c, d, by
the formula, give a number of combinations equal to 2^ — 1 ^ 16 — 1
8=3 15; and we, in fact, have a, b, c, d; ah, ac, ad, be, bd, cd; abc, abd,
acdy bed; and abed,
401. We will now deduce an analogous formula to the above, for the
sum of the permutations of n letters, taken 1 and 1, 2 and 2, 3 and 3,
n and n, when each letter is combined with itself, so as to be taken to
the second^ third, fourth, and . . . n^^ powers.
PERMUTATIONS IN WHICH THE LETTERS ARE REPEATED.
Two letters, a and 5, permuted in this way, give the four permuta-
tions, aa, ab, ba, bb. Hence, the number of permutations of two let-
ters, taken 2 and 2, when each letter is associated with itself, is equal
to 2^ Three letters, a, b, c, give the nine permutations, aa, ab, ac, ba,
cay bb, be, cb, cc. Hence, the number of permutations of three letters,
taken 2 and 2, is equal to 3'. Four letters, a, b, c, d, give, when taken
2 and 2, the sixteen permutations, aa, ab, ac, ady ba, ca, da, bhy bcy
bd, eb, db, cc, cd, dc, dd. Hence, the number of permutations of four
letters, taken 2 and 2, is equal to 4*. And, in general, it is plain that
the number of permutations of n letters, taken 2 and 2, is equal to n*.
We will next show that the number of permutations of » letters, taken
3 and 3, is equal to n\ The three letters, a, b, and c, when permuted
in this way, give the twenty-seven permutations,
aaa, aab, aac, aba, baa, oca, caa, ahc, cba;
bbb, bbc, bba, bcb, ebb, bah, abb, hac, hca ;
ccc, ccb, cca, cbc, bee, cac, ace, cab, ach.
Hence, the number of permutations of three letters, taken 3 and B ,
is equal to 3^ And, in general, of n letters, taken 3 and 3, the rnxxi:^-
ber is n^ In like manner, the number of permutations of n ^®**'^^>
taken 4 and 4, is expressed by n\ Hence, for the entire sum ^\^*^^
permutations, takerf 1 and 1, 2 and 2, 3 and 3 .... n and n, we
33 z
886 PERMUTATIONS AND COMBINATIONS.
(n^ —" X)n
n + n* + n* + n» = ^^ = — , the whole constitatiDg a geo-
metrical series, whose common ratio is n. Thus^ to apply the fomnila,
suppose it he required to determine the number of permutations of fire
fn- — 1> (5* — 1)5 15620 ^^^^ ^
lettenL then n = 5, and ^^ ^ = ^^ j—^ = — r- = 3905, and
« — 1 4 4
/jiB Y)n
if n = 24, the number of letters in the alphabet, we have ^ =^ =
^'^^l:rJ)^ = 1391724288887252999425128493402200.
Thus, let it be required to determine the entire number of cbang^
that can be made with the three vowels, a, e, and i. The formula
^^ :J-, becomes ^ — it— ^ = --r- . 3 = 39, and we find that we have
71. — 1 A L
the thirty-nine permutations.
a, e, f , taken 1 and 1 ; ae, at, ea, ta, ei, tV, aa, ee, tV, taken 2 and 2 ;
aeiy eaif iea, eia, iae, ate, taken 3 and 3, in the usual way ; aaaj aacy
aaiy eaOy iaa, aea, am, eee, eea, cet, aee, tcr, cae, «V, iiV, iVa, ttV, aiV,
nV, ta?', tci, taken 3 and 3, when the letters are combined with them-
selves.
PARTIAL PERMUTATION.
402. Sometimes the nature of the problem is such, that some of the
quantities cannot be permuted with each other. Thus, let it be re-
quired to determine how many words of two letters each can be formed
out of the letters a, e, c, d, admitting that the consonants afisodated
together will not form a word. The formula for the number of per-
mutations of m letters, taken n and n, gives us 4 (4 — 1) = 12. But
we have to reject the number of permutations of two letters, c and </,
taken 2 and 2, or 2 (2 — 1) « 2. Hence, 12 — 2 a: 10, the number
of permutations that can be formed in the required manner. Thej
are, ar, ac, ady ea, cOy da, ec, edy cCy de.
RULE.
Find the entire number of pertntttations of n letterSy taken n and d,
and from (his iubtract tJie numher of permutatians o/ the p eqtectal
Ir/frrSj taken n and n.
OBNEBAL EXAMPLES IN PERMUTATIONS; ETC. 887
EXAMPLES.
1. How many words of two letters each can be formed out of the 4
first letters of the alphabet^ assunuDg that oonBonants alone will not
form a word ? ^
Ans. 4 (4 — 1) — 3 (3 — 1) t= 6 ; words, ah, ac, ad, ha, ca, da.
2. How many numbers composed of 3 digits each can be formed
from the number 24371, in such a way that none of the numbers con-
tains only odd digits?
Ans, 5(5 — 1) (5 — 2) — 3(3 — 1) (3 — 2) = 54. Numbers,
243, 234, 342, 324, 432, 423 ; 247, 274, 472, 427, 742, 724 ; 241,
214, 412, 421, 142, 124 ; 237, 273, 372, 327, 732, 723 ; 231, 213,
312, 821, 132, 123 ; 271, 217, 721, 712, 172, 127 ', 437, 473, 374,
347, 734, 743 ; 431, 413, 341, 314, 134, 143 ; 471, 417, 741, 714,
174, 147.
3. How many words of 6 letters each can be formed out of 6 voweh
and 6 consonants, assuming that ihe consonants by themselves cannot
form a word?
Ans, 12(12 — 1) (12 — 2) (12 — 3) (12—4) (12 — 5) —
6 (6 — 1) (6 — 2) (6 — 3) (6 — 4) (6 — 5) » 664560.
4. How many words of six letters each cai^ be formed from the 20
consonants and 6 vowels of the alphabet, assuming that the consonants
alone cannot form a word ? Ans. 137858400.
GENERAL EXAMPLES IN PERMUTATIONS AND COMBI-
NATIONS.
1. Three travellers on a journey reach three roads: in how many
different positions, with respect to each other,^ may they oontinii6
travelling, provided, that no two of them pursue the same road ?
Ans. 6. A may take the first road, B the second, and C the
third ; or, A the first, C the second, B the third, &o.
2. How many days can 3 persons be placed in a different position
at dinner ? Arts. 6 days.
3. How many days can 8 persons be placed in a different position at
dinner ? • Ans. 40820 days.
388 QENEBAL EXAMPLES IN
4. In how many different positions can a staged driver place the 4
horses of his team in harness ? An$, 24.
»
5. How many changes may he made in the words of the sentence :
" Lazy hoys make worthless, vicious men ?" Ans. 720.
6. How many changes may he struck with 10 keys of a piano ?
Ans. 3628800.
7. How many words of five letters each can he made out of the first
24 letters of the alphabet, assuming that no letter is repeated more
than once in the same word ? Ans, 5100480.
8. How many words of three letters each can be made out of the 24
first letters of the alphabet, with the same proviso as the preceding ?
Ans. 12144.
9. How many numbers of two digits each can be made out of the
numbers 1243 (provided, that none of the digits of 1243 appear twice
in the same number), and what are they ?
Ans. 12. Numbers, 12, 43, 14, 13, 21, 84, 41, 31, 24, 23, 42, 32.
10. How many numbers of five digits each can be formed out of the
number 1876548, provided, that no digit of the given number appear
twice in the required results ? Ans. 2520.
11. How many changes can be made in eveiy file of 2 men in a
squad of 20 men, so that the files shall differ by at least one man Y
A71S. 190.
12. How many changes can be made in the position of the first 12
letters of the alphabet ? Ans. 479001600.
13. How many numbers of 1, 2, and 3 digits can be formed out of
the number 476, provided, that no digit is repeated in the same
number ?
Ans. 15. Numbers, 4, 7, 6 : 47^^ 46, 74, 64, 67, 76 ; 476, 467,
746, 764, 647, 674
14. How many numbers of 1, 2, and 3 digits can be formed out of
the number 476, when each digit is repeated 1, 2, and 3 times in the
respective numbers ?
Ans. 39. Numbers, 4, 7, 6, 44, 47, 46, 74, 64, 67, 76, 77, 66, 476,
467, 746, 764, 647, 674 ; 444, 447, 446, 744, 644, 474, 464 ; 777,
774, 776, 477, 677, 747, 767; 666, 664, 667, 466, 766, 646, 676.
PERMUTATIONS AND COMBINATIONS. 380
15. How many ikimberR of 1, 2, 3, 4, and 5 digits can be formed
out of the number 56789, provided, that no digit occurs more than
once in the connection with exactly the same digits ?
Ans. 31 . Numbers, 5, 6, 7, 8, 9 ; 56, 57, 58, 59, 67, 68, 69, 78,
79, 89 ; 567, 568, 569, 578, 579, 589, 678, 679, 689, 789 ; 5678,
5679, 5689, 5789, 6789 ; 56789.
16. Three travellers, A, B, C, on their journey come to
three roads, and may either travel aU together on one of the
three, or singly, on the three roads, or two on the same road,
and the third on a different road. How many selections may
they make of their routes ?
Ans. 27. Three, when all together; six, when they go separately; six,
when A and B go together, and C goes by himself; six, when A and
G are together, and B by himself; six, when B and C are together,
and A by himself.
17. How many numbers composed of three digits each can be made
out of the number 123456789, provided, that all the numbers thus
formed shall differ by at least one digit. Ans. 84.
18. How many numbers of 4 digits each can be formed out of the
same number, 123456789, so as to fulfil the above condition ?
Ans. 126.
19. How many numbers composed of single digits, of 2 digits, of 3
digits, of 4 digits, and of 5 digits, can be formed out of the number
12345, in such a way that each number shall differ from all the other
numbers by at least one digit.
Ans. 31. Numbers, 1, 2, 3, 4, 5 ; 12, 13, 14, 15, 23, 24, 25, 34,
35, 45 ; 123, 124, 125, 134, 135, 145, 234, 235, 245, 345 ; 1234,
1235, 1345, 1452, 2345 ; 12345
20. How many numbers composed of 1, 2, 3, 4, and 5 digits can be
formed out of the number 12345, in such a way that the same digit
may be repeated once, twice, &c., in the same number, and the several
numbers nece^arily differing only in the position of the digits.
Ans. 3905. Numbers, 1, 2, 3, 4, 5 ; 12, 13, 14, 15, 21, 31, 41,
51, 23, 24, 25, 32, 42, 52, 34, 35, 43, 53, 45, 54, 11, 22, 33, 44,
55; 111, 112, 113, 114, 115, 211, 311, 411, 511, 121, 131, 141,
151, &c.
33*
390 LOGARITHMS.
21. A gentleman being aaked what he would take for a valuable
horsei replied, that his price was a cent for erery change that he conld
make in the position of the 82 nails in the horse's shoes. What was
his price?
Am, 2631308369336935301672180121600000 doUars.
LOGARITHMS.
403. The logarithm of a quantity is the exponent of the power to
which it is necessary to raise an invariable quantity, called the hcae^ to
produce the quantity.
Let a be the invariable quantity or base, x its logarithm, and y the
quantity given ; then, a' = y. In this equation, x is the logarithm
of y. It is usual to write logarithm, log., or simply, 1. Hence,
X ■» log^y, or 1. y. We will begin the discussion by supposing a ^ 1,
and X positive, and, though the general principles of logarithms are tme
for quantities as well as for numbers, yet, as in practice the logarithms
of algebraic quantities are seldom used, we will first show the relations
between numbers and their logarithms. Resuming the equation, a* = y^
we see that when y = 1, x = 0, for, a" = 1. Hence, log. 1 = 0, and,
since this will be so, whatever may be the value of a, we see that the
log. 1 is always zero. Every value given to x, above zero, will eause
y to increase mote and more ; when x = 1, y = a. When x has valnes
attributed to it greater than 1, y will exceed a m(Mre and more, until
X = 00 gives also y = co . Suppose, for instance, a = 8, then
X = and x = 1 give y = 1 and y = 8. Moreover, x = 1, 2, 8,
4, &c., gives y «= 8, 64, 512, 4096, &c. Finally, x = oo gives y ss ao ,
for, 8* ss 00 . It is plain that, when x has an intermediate valve
between and 1, that y will have a value between 1 and 8,
and that by making x pass through all possible positive values,
entire and fractional, between and 1, and 1 and oo , y will be made
to pass through all possible positive values between 1 and 8, and
8 and oo . The contrary of what we have just seen will be the case
when X is negative. For the equation, ar*=:y^ will become — = y; and
it is plain that, as x increases y will decrease, and that when x = oo ,
y = 0. To illustrate this, let a = 8, and let x = 0, 1, 2, 3, 4, ftc..
LOOARITHMB. 891
then, y = 1, ^, ^, ^j^, ,^"^5, &o. And it is evident that, by giving
to X all possible positive valaes, entire and fractional, between and 00 ,
y may be made to pass through all possible positive values between 1
and 0. Suppose now, a <^ 1. It wiU still be true that all possible
values may be formed from the powers of one number, but the order of
the numbers will be reversed. For values of x, between the limits
and — 00 , we will get all possible positive numbers between 1
and + 00 3 and for all values of x, between and + 00 , we will get
all possible positive numbers between 1 and 0.
The remarkable fact that all positive numbers might be regarded as
the powers of one invariable number, led to the invention of logarithmic
tables by Napier, for the purpose of abridging complicated numerioal
calculations. The invariable number is called the base of tJie system of
logarithms, and may evidently be any number whatever, except unity.
Since all the powers of unity are unity, it is plain that this number
cannot be taken as a base. The base of the common system of loga>
rithms is 10. The powers of this number are more readily formed
than the powers of any other number whatever, and this was the main
reason for its selection.
We have (10)° == 1, Hence, log. 1 =0,
(10)' = 10, " log. 10 =1,
(10)«=100, " log. 100 =2,
(10)» = 1000, « log. 1000 = 3,
&c. &c.
We see that, as the number changes (the base being the same), the
logarithm also changes. It is plain, moreover, that any change in the
base, the number being' unaltered, will produce a change in the loga-
rithm. Thus, if the base is 12, since 12' = 12, we have log. 12 = 1.
The logarithm of 10 in this system will then not be 1, as in the system
whose base is 10, but will be some fractional number less than 1.
Hence, we see that logarithms depend upon the number and upon the
base. That part of the logarithm in any system which depends upon
the base is called the modulus of the system. A table 0/ logarithms is a
table exhibiting the logarithms of all numbers between certain limits,
as and 100, or and 10000, calculated to a particular base.
The general properties of logarithms are independent of any particular
base. A few of these general properties will now be demonstrated.
892 LOGARITHMS.
First Property.
404. The logarithm of the product of any number of fiictors taken
in the same system, is equal to the sum of the logarithms of thoee
^tors.
For, let a represent the base of the system, y, y, y, y"', &c., several
numbers, and x, x', a/', xf'', &c., their corresponding logarithms. Then,
a' = y, a'' = }/j a*" = y", a* " = y"', &c. Multiplying these equa-
tions together, member by member, we have /:i«+«+«"+«"'+*e. =yyy'y,
&o. Then, since the exponent of the base is (from the definition) the
logarithm of the second member, we have x + a/ + x" + a/" + Ac.
= log- yyyy^ &c. But the first equations give x = log. y, ac' =r
1<^. y, a:" = log. y", &c. Hence, log. y + log. y + log. y" + log y"'
+ &c. = log 2/2/2/'^" &c., as enunciated. Then, since the logarithm
of the product of any number of factors is equal to the sum of the
logarithms of these factors, it follows that to multiply by means of l(^a-
rithms, we have only to add together the logarithms of the factors, and
to find the number corresponding to this sum. This number vrill be
the product required. Thus, let it be required to multiply 100 by 10,
by means of logarithms. The log of 100 is 2, in a system whose base
is 10 ; the log of 10 is 1 ; the sum of these logarithms, 3, is the loga-
rithm of the product. Look for the number corresponding to 3 as a
logarithm, and you will have 1000 as the product required. To exhibit
the whole in an equation, we have, log. 100 x 10 = log. 100 + log. 10
= 2 + 1 = 3; the number corresponding to which is 1000.
We have, then, for multiplication by means of logarithms, this
RULE.
Add together the logarithms of the several /actors, Thit sum toiU he
equal to the logarithm of the product. Look in the table of logarithms
for the number corresponding to this logarithm^ and t/ou voUl have ike
product required,
EXAMPLES.
1. Required the equivalent expression to log. abed,
Ans. Log. a + log. b -f log. c + log. d,
2. Kequired the equivalent expression to log. 6an.
Ans, Log. 6 -f log. a + Ic^. n.
LOGARITHMS. 398
3. Bequired tke equivalent expression to log. (a + by c* <£'.
Ails, Log. (a + by + log. c"" + log. d».
Sexu)nd Property,
405. The logarithm of the quotient arising from dividing one quan-
tity by another is equal to the logarithm of the dividend minus the
logarithm of the divisor. For^ let x equal the quotient of two quan-
AM
tities, m and n, then x= — , or nx = m. And, sinoe, when two
n
quantities are equal their logarithms must be equal, we have log. nx
=s log. m, or log. n + log. x = log. m, Henoe, log. x = Ic^. m — log.
n, as enunciated. From this we derive for division, by means of loga-
rithms, the following
RULE.
From the logarithm of the dividend subtract the logarithm of the
divisor; the remainder will be the logarithm of the quotient. The
number in the table corresponding to this logarithm toiU be the quotient
required.
Thus, to find the quotient of 1000 by 10, we have log. 1000 — log. 10
= 3 — 1=2, and the number corresponding to 2 as a logarithm is
100.
EXAMPLES.
1. Find the equivalent expression to log. -=— .
on
Ans, Log. am — log. &w, or log. a -f log. m — log. &-~log. «.
2. Find the equivalent expression to log. ' ^ .
Ans, Log. X -f log. y — log. (b -|- »).
X I fl
3. Find the equivalent expression to log. 7^^ .
Ans, Log. (x-\-y) — log. b — log. «.
xv
4. Find the equivalent expression to log. ^,
Ans, Log. X -j- log. y — log. b — log. n.
894 LOOARITHMS.
5. Find the equivalent expression to log. (^)-
An$, Log. (xt/y — log. (6n)».
6. Find the equivalent expression to log. ( 1.
Ans, Log. (a + i)— log. cl
Thtrd Property J
406. The logarithm of the power of a number is equal to the expo-
nent of the power into the logarithm of the number.
For, take the equation, a< = y, and raise both members to the m*^
power, we will have a»* = y^ (A). In equation (A), m is any number
whatever, positive or negative, entire or fractional. But, since the
exponent of the base is the logarithm of the second member, equation
(A) gives mx = log. y" ; and the first equation gives x =s log. y.
Hence, m log. y = log. y", or log y» = w log. y, as enunciated. It
follows, then, that to raise a number to a power by means of logarithms,
we have only to apply this
RULE.
Multiply ike logarithm of the numher, az found in the table, by tke
exponent of the power to which it is to be raited. This product wtU he
the logarithm of the power ^ and. the number found in the table corres-
ponding to this logarithm will be the required power.
Thus, to raise 10 to the second power by means of logarithms, we
multiply the logarithm of 10, which is 1, by 2, the exponent of the
power. The product, 2, is the logarithm of the power, and the number
corresponding to this logarithm is 100. So that we have log. (10)* =
2 log. 10 s= 2 ; the number corresponding to which is 100.
EXAMPLES.
1. Find the equivalent expression to log. (a + 6)".
Ans. m \og. (a + h).
2. Find the equivalent expression to log. c» (a + b)\
Ans. n log. c + m log. (a + 6).
The 1st. property is here used in connection with the 8d.
IiOOARITHMS. 805
8. find the equivalent expreasion to log. a* (mr);
Am. n (log. a + log. r + log. m).
Ans. n (log. a -f log. b — log. c).
The 2d. and 3d. properties both employed.
5. Find the equivalent expression to log. ^^ — p— ^ .
Ans. m log. (a + h) — 2 log. A.
a* 2c«
6. Find the equivalent expression to log. r — ^3 .
(a + x)
-4»«. Log. (a — x) — 2 log. (a + x),
(a -4- ajV
7. Find the equivalent expression to log. ~ -^.
Ans, 2 log. (a + 35) — log. (a — «).
JFbttr^A Property,
407. The logarithm of the root of a number is equal to the logarithm
of the number divided by the index of the root.
For, take the equation, a» = y, and extract the m*** root of both
members. Then, v'a' = Vi/, or, am z=z^i/. Hence, — =log. Vy
(A.), the exponent of the base being the logarithm of the second
member. But the first equation gives x = log. 1/ ; equation (A) then
y y
becomes log. — = log. Vy, or log. V^y = log. — , as enunciated.
For extracting roots by means of logarithms we have, therefore, this
RULE.
Divide the logarithm of the number, whose root is to be extracted, by
the index of the root. The quotient will be the logarithm of the root;
the number found in the table, corresponding to this logarithm, wiU be
the root required.
Thus, to extract the square root of 100 by means of logarithms, we
divide 2, the logarithm of 100, by 2, the index of the root. The quo-
896 LOGARITHMS.
tient, 1, is the logarithm of the root, and 10, the Dumber corresponding
to this logarithm, is the root required.
EXAMPLES.
1. Find the equivalent 'expression to log. ^x + y,
n
2. Find the equivalent expression to log. \/a* — x*.
. Log. (a* — a^) log. (a -\- x) (a — a:) , ,
Ans. -^"-^2 ■ - ./ = i log- (« -f a-)
+ } log. (a — xy
3. Find the equivalent expression to log. a' ^ a*.
Ant. -^ log. a.
4. Find the equivalent expression to log. y (a + by,
Ans. I log. (a + h).
408. Since, a° = 1, it matters not what is the value of a, and, since the
equation, a° = 1, gives = log. 1, it follows that the logarithm of unity
in every system of logarithms is equal to zero. From this we readih^
deduce the
Fifth Property.
The logarithm of the reciprocal of a number is equal to the loga-
rithm of the number taken negatively.
For, let N be the number, then -^ will be the reciprocal of the num-
ber, and we have log. -^ = log. 1 — log. N = — log. N «» — log. N,
as enunciated.
EXAMPLES.
(a -h xY
1. Required the equivalent expressions to log. ^^ — "^ , and log.
Ans. 2 log. (a + x) — 3 log. cc, and 3 log. x — 2 log. (a + x).
LOGARITHMS. 397
f^a X*)
2. Required equivalent expressions to log. y -^, and log.
(« + xY
a« — X** ^
Ans. Log.(a — x) — Slog. (a+a;),and31og. (a+x) — log.(a — x).
CoroUary,
409. When o ^ 1, we have a'^ss co , Hence, oo = log. oo . That
iS; the logarithm of infinity in a system, whose base is greater than unity,
is, itself, infinity. And, since = — , it follows from the fifth property,
that log. =s — 00 . That is, the l(^rithm of zero in a system, whose
base is greater than unity, is minus infinity.
When a <^ -j^j^, a number, j\j for example, it is plain that the base
must be raised to a power denoted by — oc, in order to produce in-
finity. Thus, Ct'u)""* = } I >>«> = (10) • = 00 . Hence, the logarithm
of infinity in a system, whose base is less than unity, is minus infinity.
Hence, by the fifth property, the logarithm of zero in a system whose
base is less than 1, is plus infinity. Thus, it is plain that (j^;)* s= 0,
from which oo ss log. 0.
Sixth Property.
410. The logarithm of any base, taken in its own system, is unity.
For, suppose there are any number of bases, a, 5, c, &c., and desig-
nate the logarithms in the systems, of which these are the bases, by
log. log.' log." &c. We have a' =^ a^V ^=^ ft, c' = c, &o. Hence,
1 ass log. a, 1 = log.' 6, 1 = log.' c, &c.
Thus, 1 is the logarithm of 10 in a system whose base is 10. But,
in the system whose base is 8, the logarithm of 10 must be greater
than 1. Because, 8* = 8, or 1 = log. 8. Hence, a number greater
than 1 must be the logarithm of 10.
Seventh Principle,
411. If we have a table of logarithms calculated to a particular base,
tbe logarithms of the numbers in this table, divided by the logarithm
of a second base, taken in the first system, will give, as quotients, the
logarithms of the same numbers in the second system.
34
398 LOGARITHMS.
For, let a* = f/j and h* = y. Then a* = I* (A). Distingaish Uie
logaritlims of the two systems by log. and log/. Taking the log. of both
members of (A), we have x log. a = z log. h. But, since x = log. jr,
and z = log', y, and log. a = 1, this equation becomes log. y a= log.'y,
lofiT. V
log. 5. Then log.' y = ^ — '-J-, as enunciated.
412. The two systems in most common use are the Napierian, whoee
base is 2.718281828, and the common, whose base is 10. Suppose
the Napierian logarithms to be designated by log.', and the common by
log. Then, if we knew the common logarithm of any number, we can
get the Napierian logarithm of the same number, by dividing the
common logarithm of the number by the common logarithm of the
Napierian base; or we can get the common logarithm of a number,
knowing its Napierian logarithm, by multiplying the Napierian logarithm
by the common logarithm of the Napierian base.
DIFFERENTIAL OP a*.
413. The Binomial Formula enables us to find the dilSerential of a'.
For, let u ==: a', and give x an increment, A. Then, «' 3= a*+* a= d'a^,
and tt' — u = a'^a^ — 1). Let a = 1 + 5, and develop a^ bj the
binomial formula, we will have a* = (1 + ft)** =s 1 -f A6 + -^ — ^^
+ 1.2.3 +*<'• = 1 + n^- 2 +T- 4 +*4
by neglecting the higher powers of A, as infinitely small quantities ot
the second order and higher orders. But, since 6 = a — 1, we have
«<> = 1 + h((a-\)-^-^^^ + ^^—^ -&c.) » 1 + NA. bj
designating the quantity within the parenthesis by N. Hence, u' — %
= d'(a^ — 1) = a'N/i, and du =s o*N(7j7 = v'Sdx', from which
1 J.,
c?a: = X7 • — • And, since « = a', we have x = log. ti, and dx =«
N tt
d(\oQ. u). Moreover, since a is the base, and N depends entirely upon
a, we see that the differential of a logarithm is equal to the reciprooal
of a constant, dependent upon the base into the differential of t^
quantity divided by the quantity.
LOGABITHMS.
EXAMPLES.
1 Required differential of h^. Ans. Nh^dy.
2. Required differential of log. (1 + x).
1 d(l+x) 1 dx
N* 1 + x N'l + a;'
3. Required the differential of log. (1 + a;)'.
. 1^ 2( 1 4- x)dx _ 2 dx
N* (1 + x)'* ■" N'(l+x)-
LOGARITHMIC SERIES.
414. A loearithmic series is one which will enable us to calculate
approximativelj the logarithms of any number whatever. We have seen
that the logarithm of a number depends^ Ist, on the number ; 2dy on
the base. Hence, the development of a logarithm must contain the
number, or some quantity dependent upon the number, and some
quantity dependent upon the base. These two facts serve as a guide
in assuming the form of development. Suppose we have the equation,
a^ =. x, in which x is the number, and y the logarithm. Let us
assume log. x = A + A':c -f A"jc' + A"V -f &c., in which A, A',
A'', &c., are independent of x, and dependent upon the base, a. Now,
if we make a; = 0, the first member becomes infinite, whilst the second
reduces to a finite quantity, A. The assumed form is then wrong.
Again, assume log. x = Ax 4- A'x + A"cc* + &c. Making a: = 0,
we get =fc 00 = 0, which is absurd. The development then cannot be
made under the second form. An examination of the two forms shows
that we have assumed the logarithm to be developed in the powers of
the number, and we have seen that it cannot be so developed. Since
log. x cannot be directly developed, let us see whether log. (1 + x) can.
be expanded into a series. Assume log. (1 + x) = Ax + A'x" -+-
AV + A'"x* + &c. (B). When x = we have log. 1 = 0, whicb
presents no absurdity. Taking the differentials of both members of (B),
and dividing out by dx, we have, -=^ . ^— — = A + 2 A'a + 3 A"x' -|-
4A'V + &c. (C), in which, N == (a — 1)^1—^ + g
>« &c. Making X = in (C), we get A = 1. Differentiating (G),
400 LOGARITHMS.
we get, after dividing out by dx, — :j^ y ^ v g= 2A' + 2 . 3A"ar +
3 . 4 A'V + &c. (D), whence A' = — ^^. Differentiating (D), and
dividing out by dx, we have + |= . ^ ^^ = 2 . 3A"+2 . 3 . 4A'^x
-f &c. (E). Whence, by making a: = 0, there results, A" = + -^.
The law for finding the coefficients is now plain ; K!" will be found equal
1 +1 1
to — -r^^, A>-^ = -^xT-> A^ = — ^r^y &o. Replacing the constants
4N oN 6N ^
1 / 0^0?
in (B) by their values, we have log. (1 + a:) = ^ I x — "9 + "o" —
|-V|'-&o.). (P.)
Since (1 + ^); the number whose logarithm is developed, has been
expanded in the powers of x ; we conclude that, though the logarithm
of a number cannot be developed in the powers of the number, it may
be in the powers of a number less by unity.
Since N = (a — 1) — ^^ — ~ — ^ + ^ — 5 — -y &o., in which a is the
base of the system, it is plain that the factor, t^, in equation (P), depends
for its value upon the base alone. Equation (P) shows, moreover,
that the logarithm of a number is composed of two factors, one
dependent upon the base alone, and the other dependent upon the
number alone. The factor which depends upon the base for its value
is called the modulus of the system. The modulus is usually represented
by M, or M'.
415. If we take the logarithm of (1 + x), in a system whose base
is a', and denote the new logarithm by log.', it is plain that the new
development will differ from the first only in this notation and in the
modulus. Hence, we will have
log. (l+x) = m(x — -^ +y— -J +-g-— &c.), and
log.'(l + x) = M'(x-^+|'-J* + |'-&c.).
Hence, log. (1 -f- x) : log/ (1 + x) : : M : M'.
That is, the logarithms of the same number in two different Sjystems
are to each other as the moduli of these systems.
LOOABITHM8. 401
416. Sinoe the modulus is expressed in tenns of the base, it follows
that if the base be giyen, the modulus can be determined, and con-
yerselj, the base can be found from the modulus.
If, then, we make M' = 1, the base of the system wiU become knowv
The system of logarithms, whose modulus is unity, is called Napierian,
from Lord Napier, the inventor of logarithms. The base of the system
calculated from the assumed modulus, is 2*718281828.
COMMON AND OTHER LOGARITHMS FOUND FROM
NAPIERIAN.
417. The proportion, log. (1 + x) : log'. (1 + a:) : : M : M', be-
comes, when M' = l, log. (1 -f cr) : log'. (1 + a:) : M : 1. Hence, we
have log. (1 + a;) = M Ic^. (1 + x) (A).
This equation shows that the Napierian logarithm of any nwnber,
multiplied hy the modulus of any other system, wiU give the logarithm
of the same number in that system. Moreover j equation (A), compared
vfith the 7th Property, shows that the modulus of any system, is equal to
the logarithm of the Napierian base of that system,
MEASURE OF ANY MODULUS.
418. Let a be the base of any system, and let the logarithms of this
system be designated by log., and thos^ of the Napierian by log'. Let
(1 H- x) be any number, then the 7th Property will give log. (1 H- »)
= . — r- Jos'- (1 + ») (B). Equation (B) compared with (A) shows
that M = - — 3 — . Hence, the measure of the modulus of any system
log. a
is equal to the reciprocal of the Napierian logarithm of the base of that
system.
This measure is even true for the Napiarian modulus itself, for we
have M' = r— j — = =- = 1.
log . € 1
TABLE OF NAPIERIAN LOGARITHMS.
419.Theformula,?(l+a;) = M(a;— ^+^— ^+-5— e + *®->>
34* 2a
402 LOGARITHMS.
a:" a:* X* rc* x*
becomes, wben M = 1, r(l + x) = (x — — + ,^ r4--R
^ O 4: O
+ &c.,) (C).
Making x = 1, we have, /'2 = 1 — } + i — i + J — l+&c.
6
This series enables as to calculate the Napierian logarithm of 2, and,
by making a; r= 2, 8, 4, 5, &c., we can calculate the Napierian logarithin
of 3, 4, 5, 6, &c. But the series does not converge rapidly enongli,
because, for every number greater than 2, the series goes on increasing
continually, and it would be necessary to take an infinite number of
terms in order to make the calculation in any degree correct. A simple
artifice will enable us to convert the above series into a converging one,
that is, into a series in which the terms will become smaller and smaller ;
so that all after a certain number may be neglected without affecting
the result materially. Two transformations are used to affect this con-
vergence.
First Transformation,
420. Make x = - in equation (C), it will become
- ^ + &c., (D).
This series will become more converging as y increases.
Making y = 1, 2, 3, 4, 5, 6, &c., in succession, we get
n—n=r2=i—i + i_ j + ^_j + &c.,
i'4 = rs + i-j^ + jV - jh + t2't» - &«•,
r5 = n + i — ,>3 + ji^ — 7,'„ + jjj5 - &c.
The first series enables us to calculate the Napierian logarithm of 2.
The second series enables us to calculate the Napierian logarithm of 3
when that of 2 is known ; and the formula, evidently, only can be nsed
when the logarithm of the next lowest number is known.
Second Transformation.
421. The preceding formula does not give a sufficiently converging
series for small numbers. A better series can be obtained in the fol-
LOGARITHMS. 403
x' 01? tl^
lowing manner. In the equation, ^(1 4-a;) = x — -5-4- -^ -r- +
A o 'x
-J- -r + &c., we get, by changing + x into — x, ?(1 — a;) = — x
a;* x' X* cc* a;'
— *ir"^"Q" — T ^ — "fi" — ^®* Subtracting the second series
from the first, we get V{l + x) — V(l'^x^ = vU-^\z=2(x +
T+T + T + ? + fl+S+**^-')(^)- Pl'cc^=l +
-, in which z is a whole number. Then, x = 5 =-, and replacing x
in (B) by this value, we get ^(1 + -), or ^(1 + a) — V z—2 U — ^
+ r^rn)' + 572^' + (T2rfiy + H This series con-
verges more rapidly than (D), and will answer even for small numbers.
Let 2 = 1, 2, 8, 4, 5, 6, &c., we will have
?8-r2-2(i + g^ + g^ + ^-^ + -L^ + &c.
__/ 1 1 1 1 1
f6 — 75 = 2 ^Tj + gp-j + g^jjy. +7-(ii)J+ 9(n)»'^ 11(11)''
+ ^0
»
Remarks,
1. For small numbers it is necessary to take a good many terms of
this series, but when the numbers are large one or two terms will
answer. Thus, let x = 1000, then Z'1001—nOOO= 2(2^^1 + 3?2()0ly
4. + &cV We see that the first term of this series will
5(2001)* /
only add about -j^^j^ to the logarithm of 1000, and the second term
will add less than ffuUTFTJ^UiJTJO- ^^^^ 8^°^® logarithms are seldom
404 LOGARITHMS.
carried further than the 7th place of decimals^ all the terms after the
first may be neglected.
2. In constructing a table it is only necessary to calculate the loga-
rithms of prime numbers, since the logarithm of any number made up
of factors is equal to the sum of the logarithms of its factors. Thus,
no = r2 + r5, v\2 = rs + r4 = rs + 2^2, &c.
3. Knowing the Napierian logarithm of 10, we can find the modulus
of the common system. For, log.' 10 : log. 10 : : 1 : M. Hence,
M = y' ; but log. 10 = 1, and the calculated value of log/ 10 is
2.302585093. Therefore, M = 2.302585093 "* ^'^^4294482.
If we now multiply the Napierian logarithms found from the series
by this modulus, we will have a table of common logarithms.
ADVANTAGES OF THE COMMON SYSTEM.
422. 1. A fixed law for the characteristic.
The characteristic of a logarithm is its entire part. Thus, 2 is the
characteristic in the logarithm 2-302585093.
Whatever may be the base of the system, the logarithms of the
powers of that base will contain no decimals, and, therefore, be charac-
teristics only. Thus, let 7 be the base, then the log. 7=1,
log. 49 = 2, &c.
In the common system, with the base 10, we have log. 10 = 1,
log. 100 = 2, log. 1000 = 3, &c., and we see that the characteristic is
always one less than the number of figure places in the number. This
is true, whether the number be an exact power of the base or not ;
thus, 512, lying between 100, whose logarithm is 2, and 1000, wboee
logarithm is 3, has for its logarithm 2 and a certain decimal. Now,
10 is the only number in the whole range of numbers, which, taken as
a base, will give logarithms whose characteristics are formed according
to a fixed law.
In a table of common logarithms, the characteristics are not written,
since they are always one Jess than the number of fibres tn the ffiven
numbers,
2. A table of logarithms for decimals need not be constructed.
Since, Log. -1 = log. y'^ = log. 1 — log. 10 =s — 1.
Log. -01 = log. y J^ = log. 1 — log. 100 « — 2.
Log. -001 = log. y^Vu « log. 1 — log. 1000 « — 3.
LOGARITHMS. 405
And, since, in general, log. ^^^ = log. 1 — mlog. 10 =bs — m, it is
plain that tlie cbaracteristic of a decimal is negative, and one greater
than the number of ciphers between the decimal point and the first
significant figure. This is true, also, when the decimal is not the re-
ciprocal of some power of the base. ,Thus, log. -08 = log. y^ = log.
8 — log. 10 = a decimal — unity. Hence, the log. '08 must have a
negative characteristic equal to minus unity. Moreover, if we have a
decimal, such as -00278 = tdVo^u* ^® ^^^ h&Ye log. '00278 =s= log.
278 — log. 100000. And, since log. 278 is 2 whole number, plus a cer-
tain decimal ; and, since the log. 100000 is 5, the log -00278 = — 3,
whole number plus a certain decimal. We see, from this example, that the
logarithm of a decimal is obtained by regarding the decimal as a whole
number, and prefixing a negative characteristic, one greater than the num-
ber of ciphers, between the decimal point and the first significant figure.
The foregoing reasoning can be extended to any decimal whatever, be-
cause, change the decimal into an equivalent vulgar fraction, it will be
seen that the number of figure places in the denominator exceed^ those
in the numerator by one more than the number of ciphers between the de-
cimal point and the first significant figure of the given decimal. Now, if
the base were any other number whatever than 10, it is plain that a table
of logarithms for whole numbers would not be applicable to decimals.
3. The logarithm of a mixed decimal y such as 42-733, can he found
from a table o/ logarithms constructed to the base 10, b^ regarding the
mixed number as a whole number j and prefixing to the decimal part
of the logarithm found in the table a characteristic one les^ than the
number of figure places in the entire part of the mixed number.
For, log. 42-733 = log. \%\%^ = log. 42733 — log. 1000 « 4 +
a certain decimal — 3 = 1 + a decimal.
The foregoing rule for finding the logarithm of a mixed number can
evidently be applied to any mixed number whatever.
4. The logarithms of numbers which differ only in the number of
annexed ciphers, will differ only in their characteristics.
Thus, the logarithms of 126, 1250, 12500, and 125000, differ only
in their characteristics. For, log. 1250 = log. 10 -f log. 125 = 1 -f
log. 125 ; log. 12500 « log. 100 + log. 125=2 +log. 125 ; log. 125000
» log. 1000 + log. 125 = 3 4- log. 125.
This property of common logarithms is very important, since it saves
the trouble (as will be seen more fully hereafter) of constructing a table
for numbers whose figure places exceed four.
406
LOGARITHMS
APPLICATION OF COMMON LOGARITHMS.
423. The base of the common system being 10, the characteristic or
entire part of the logarithms need not be written in the tables, since, ajs
we have seen, it is always one loss than the number of figure places in
the given number. The decimal parts of the logarithms are then only
written in the tables, and are generally carried as far as six places.
The first vertical column on the left in the tables, contains the given
numbers ; the last vertical column, marked D, contains the tabular dif-
ference corresponding to two given logarithms, and is constructed by a
method hereafter to be explained. The intermediate vertical columns,
marked at top, 0, 1, 2, 3, &c., contain the decimal parts of logarithms.
The following table exhibits the logarithms and numbers between 100
and 130. When the given number contains four figure places, the
100
1
2
3
4
6
6
7
8
D
000000
0434
0868
1301
1734
2166
2698
3029
3461
3891
4.32
101
•4321
4751
6181
6609
6038
6466
6894
7321
774S
8174
428
102
srxK)
9026
9461
9876
•300
•7i4
1147
1570
1993
•J416
424
103
012887
3269
3080
4100
4521
4940
5568
6779
6197
64 16
419
104
7033
7461
78G8
8284
8700
9116
9532
9947
•361
-775
416
105
021189
1603
2016
2428
2841
3-2r.2
3C64
4075
4486
4896
412
106
6306
6715
6125
6533
C942
7:150
7757
8164
8671
8f»78
40S
107
9384
9789
•095
•600
1004
1408
If 12
2216
2619
,':021
404
108
033424
3826
4227
4C2S
6029
54;)0
6830
G2a0
&VJ9
70-28
400
109
7426
7825
8223
8C20
9017
9414
9811
•2U7
•C02
•998
396
110
041393
1787
2182
2676
29(.9
3362
3755
4148
4640
4932
393
111
5323
6714
6105
6495
6885
7276
•HUCA
8053
8442
8^30
389
112
9218
9606
9993
•380
.766
1153
1538
1924
2309
2694
380
113
063078
8463
3846
4230
4613
4996
n378
57C0
6142
C5*24
.382
114
6905
7286
7666
mw
8426
88(16
9185
9663
9942
•330
379
116
060698
1075
1462
1829
2206
25^2
29r,8
3333
3709
40S3
376
116
4458
4832
6206
6580
6963
t;'J26
6f99
7071
7443
7815
372
117
8186
8657
8928
9298
9668
• -.'J-S
•407
•706
1145
1614
369
118
071882
2260
2617
2985
3352
3718
4086
4461
4iil6
5182
966
119
6547
6012
6276
6640
7004
73CS
7731
8094
8467
8819
363
120
079181
9543
9904
•2G6
•626
•W'7
1»47
1707
2067
2420
SCO
121
082785
3144
3503
3861
4219
4.' 76
4934
5291
6647
C004
357
122
G3C0
6716
7071
7426
7718
8136
8490
8846
9198
9552
855
123
9905
•258
•611
•9G3
1316
1 fir-
2(118
2370
2721
3071
361
124
093422
3772
4122
4471
4820
Si f 9
?518
6866
6215
6662
349
125
6910
7257
7f04
7951
8"2<iS
8(44
8990
9836
9CS1
•26
126
100371
0716
1069
1403
1747
2(i91
2434
2777
3119
3462
343
127
3804
4146
44S7
4828
6169
06IO
6861
6191
6631
C871
340
128
7210
7649
7888
8227
ecc6
8903
9241
9579
9916
•253
338
129
llOMiO
0026
12g;j
lf.l!l»
1934
2'-'70
2C05
2940
3275
3609
S3S
130
113943
4277
4611
4944
5278
6(511
6943
6276
f608
6940
332
first three will be found in the first vertical column on the left, and the
fourth must be sought at the top of the page, among the numbers be-
tween and 9, inclusive. Thus, 121 may be found in the first vertical
column on the left, and the last figure of 1215 will be found in the
column marked 5 at top. The logarithm of 121 is written just oppo-
site to it in the column marked 0; 1210 will obviously have the same
IiOOAR^THMS. 407
decimal part of ito logarithm, bnt a characteristic greater by unity ; the
logarithm of 1215, on the other hand, most not only exceed the loga-
rithm of 121 in characteristic, but also in decimal part, and its decimal
part is then taken from the column marked 5.
TO FIND THE LOGARITHM OF A GIVEN NUMBER BELOW 10,000.
424. When the number is less than 100, the decimal part of the
logarithm is found in the table just opposite the given number. The
characteristic will be determined by the number of figure places in the
number. For numbers above 100 and below 10,000, we have tho
following
RULE.
Look for the first three figure$ of the given number in the column
marked N, and if it contains hut three, the decimal part of its loga-
rithm mil he found just opposite these three figures in the column
marked 0. But if there are only four figures in this column, opposite
the given number, the two left hand figures above, in the same column,
where six figures are written, must be prefixed to the four figures found.
Thus, the decimal part of the logarithm of 110, is 041398, and the
entire /o^an^Am, 2 041398. The decimal part of the logarithm of
112 is 0-i921S, the first two figures being found opposite 110. The
entire logarithm is then 2049218. The logarithm of 117 is 206816,
and of 120, 2079181.
425. When the given number contains four figures, look for the first
three in the column marked N. The last four of the decimal part of
its logarithm will be found just opposite, in the column marked at top
by the fourth figure. To these four figures must be prefixed two deoi-
mala in the column marked zero, either found opposite the first three
figures, or above, where six figures occur. Thus, to get the logarithm
of 1213, we first find 121 in the first column, and then look just oppo-
site, in the column marked 8, where we find 3861, and prefix to it 08,
taken from the column marked 0. Hence, the logarithm of 1213 is
3-083861. The logarithm of 1283 is 3108227, the two left hand
figures of its decimal part being found above, where six figures occur.
When, however, we pass over a decimal with a point prefixed, the
first two figures are to be found below, and not above. Thus, the loga-
rithm of 1234 is 3091315. In like manner, the logarithm of 1231
IB 8*0902.58. The place of the point is always supplied by zero. If
408 LOGABITHMB.
the left hand figures were not taken helow, the logarithm of 1231, 1282;
kc.y would be less than the logarithm of 1280, 1229, fte.
EXAlffPLES.
1. Required the logarithm of 129. An$. 2-110590.
2. Eeqnired the logarithm of 1290. An$, 8-110590.
3. Required the logarithm of 1297. Ant. 8-112940.
4. Required the logarithm of 1285. Am. 8-108908.
5. Required the logarithm of 1238. Atu. 8-090968.
6. Required the logarithm of 1236. Ans. 3092018.
7. Required the logarithm of 119. Am. 2-0755i7.
8. Required the logarithm of 1191. Ans. 3-075912.
TO FIND THE LOGARITHM OF A NUMBER ABOVB 10,000.
426. Cut off on the right all the figures, except the first four of the
highest denomination. Find the decimal part of these four left hand
figures, as before, and prefix a characteristic, one less than the number of
figure places in the given number previous to cutting off. Multiply
the tabular difference by the figures cut off on the right, and from this
product cut off as many places for decimals as thero are figure places in
the multiplier. Add the result to the logarithm before found. Tfaos,
to find the logarithm of 129451, we cut off 51, and find the decimal
part of 1294 to be lll934. To this we prefix a characteristic, 5, and
we have 5111984, which is truly the logarithm of 129400 (Art 422).
Next, we multiply the tabular difference, 385, by 51, and cut off 85
from the product, 17085 ; we would then have 170 to add to 5*111934 ;
but, as '85 is greater than }, we increase 170 by 1, because 171 ia
nearer to 170*85 than is 170. Adding then 171, we have 5-112105
for the logarithm of 129451. In general, whenever the decimal cat
off in the product, which results from multiplying the tabular diffeieiHse
by the right hand figures of the given number, exceeds \^ the hsi
figure of the entire part of the product must be increased by 1.
The explanation of the process is simple. The tabular diile-
rence expresses the difference between the logarithms of oonsecutiye
L0QABITH1I8. 409
nnmben. Thus, Uie logarithm of 1293 is 31599 ; and of 1294 is
3*1934^ and their difference, 335^ is written in the column marked D.
This oolamn^ it is plain, will moreover express the difference between
the logarithms of nnmbers which differ by 10, 100, 1000, &c. We
have seen that the logarithm of 129400 was 5-11934, but we were
required to find the logarithm of 129451, and it remains to be seen how
much the logarithm of 129400 is to be increased. We employ the
principle, that the difference between any two numbers is to the difference
of their logarithms, as the difference between any two other numbers is
to their difference of logarithm. Hence, 129400 — 129300 : 335 : :
129451 — 129400 : x, or 100 : 335 : : 51 : x^ in which x represents
the augment to the logarithm of 129400, to give the logarithm of
129451. We find x = >f gj* = 170-85.
EXAMPLES.
1. Required the logarithm of 1309958. Am. 6117259.
In this example three places are cut off for decimals, because we took
the difference between 1309000 and 1308000. The first term of the
proportion was then 1000.
2. Required the logarithm of 13080000. Am, 7116608.
3. Required the logarithm of 13087654. Am. 7-116862.
TO FIND A NUMBER CORRESPONDING TO A GIYEN
LOGARITHM.
427. When the decimal part of the logarithm can be found in the
tables, the first three places of the required number will be found on
the left, immediately opposite this logarithm, and in the column marked
N. The fourth figure, which is to be annexed to the three already founds
is to be taken from the number of the column in which the decimal
part of the logarithm was written. Then point off, from the left of the
number thus obtained, one more place for the entire part, than is indi-
cated by the number of units in the characteristic. Thus, the number
correspondiDg to the logarithm 2107888 is 128-2, and to 1-107888 is
12*82. So, the number corresponding to the logarithm 3-088136 is
1225, and contains no decimal part.
But, when the decimal part \$annot be exactly found, take the next
less logarithm from the tables, and subt^ct this from the given loga-
35
410 LOQABITHMB.
rithm. Auuex two or mare cipbers to this difference^ and divide bj
the tabular difference. Annex the quotient to the number correspond-
ing to the next less logarithm, and you will baye the required number.
Thus, let it be required to find the number corresponding to the lo-
garithm 41132800. The next less logarithm is 41132750, and the
corresponding number 12980 ; annexing 4 ciphers to 50, the difference
of logarithms, and dividing by 335, the tabular difference, we have a
quotient, 149, which is to be annexed as a decimal to 12980. Hcnoe,
the number sought is 12980-149. It is plain that we may annex as
many ciphers as we please to the difference of logarithms. The pro-
cess just described being the reverse of the preceding, requires no ex-
planation.
EXAMPLES.
1. Required the number whose logarithm is 4*07850.
Ans. 1198118.
2. Required the number whose logarithm is 7*116608.
18080000.
8. Required the number whose logarithm is 8*075550.
An$. 119000243*85.
4. Required the number whose logarithm is 3*017249.
Am, 1040*52.
5. Required the number whose logarithm is 3*153209.
Ans, 1423*013.
Remarks,
When only one cipher has to be annexed to the difference between
the given and next less logarithm, to make it divisible by the tabular
difference, no cipher- is prefixed to the quotient. But when, (as in
example 5), two ciphers must be annexed to make the division possible,
then one cipher must be prefixed to the quotient. And when, (as in
example 3), three ciphers must be annexed, two must be prefixed to the
quotient. And, in general, the number of ciphers prefixed is one les
than the number of ciphers annexed, to make the division possible.
LOGARITHMS. 411
GENERAL EXAMPLES.
1. Required the product of 1-040, by 1-055 and 10-77.
Am. 11-81687.
For log. 1 040 = 0017033
los. 1055 = 0023252 f
log. 10-77 = 1032216
And logarithm 1072501 corresponds to 11-81687.
2. Required the quotient of 10-22 by 1016. Ans. 10-059.
For log. 10-22 = 1-009451
and log. 1-016 = 0006894
and their difiference, 1-002557, corresponds to 10-059.
3. Raise 11-48 to the third power. Aru. 1512-95.
For log. 11-48 = 1059942
3-179826 = log. 1512-95.
4. Required the thirty-second root of 1172. Ans. 1-21871.
^ . log. 1172 3068928 .__,
For we have °,,,, — = — -; — = 095904.
And the corresponding number is 1-21871.
SOLUTION OF EXPONENTIAL EQUATIONS BY MEANS OF
LOGARITHMS.
428. Suppose we have the exponential equal, a* = p, then, x log.
los^. p
a = log. p, and x = ^ ; that is, the root is equal to the logarithm
of p divided by the logarithm of a, and this root itself may be foi\nd
by means of logarithms for log. x = log. . ■ = log. log. p — Ic^.
log. a.
EXAMPLES.
1. Required the value of x in the equation, 10* = 10.
Arts, a; = 1.
For, X log. 10 = log. 10, or a; (1) = 1, or a* = 1.
412 ABITHMETIOAL PEOOEE88ION.
2. Required the yalue of x in the equation, 12* ss 13.
An*. 1032.
For, log. 18 « 1-118943
and, log. 12 » 1 079181
log. 1118948 = 0046862
log. 1079181 = 033094
0-013768.
Hence, log. x = 0-013768, or x = 1032.
3. Required the value of x in the equation, 100* a 1000.
J.f|«. X = {.
4. Required the value of x in the equation, 10* = 125.
Ans. X = 2-096911.
5. Required the value of x in the equation, 100* = 125.
Am. X = 1-048455.
6. Required the value of x in the equation, 11* = 11.
Ant, X = 1.
For, log. X =.log. log. 11 — log. log. 11 = log. 1-041393 — log.
1 041393 = 0-017614 — 0-017614 = 0. Now, since log. x = 0, x
must he unity.
The result in this case might have heen anticipated, but ire have
carried out the process to show its truthfulness.
ARITHMETICAL PROGRESSION.
429. Quantities, which increase or decrease hj a common dij
rence, are said to be in Arithmetical Progression.
Thus, 2, 4, 6, 8, 10, &c., constitute an arithmetical progression, the
common difference being 2. In like manner, 10, 8, 6, 4, 2 constitute,
also, an arithmetical progression, the common difference being 2.
When, as in the first case, the common difference is an increment, the
series is called ascending ; and when, as in the second case, the com-
mon difference is a decrement, the series is called descending. The
quantities, a^ a '\' d, a •\' 2d, a -\- 3e7, &c., constitute an ascending
series ; and a, a — d, a — 2(f, a — Zd, constitute a descending series.
ABITHMETIOAL PROGRESSION. 413
It is evident, likewise, that the quantities, a ■\' ^d, a + 2d, a -{- d, a,
form a descending series ; and, in general, an ascending series, when
read in reverse order, becomes a descending serieS; and conversely.
It is plain that the natural numbers, 1, 2, 3, 4, 5, 6, 7, &o., are in
arithmetical progression, the common difference being 1.
It is plain that, if we know any term of an ascending series, we can
find the succeeding term by adding to the known term the common
difference. And, if we wish to find the term succeeding any known
term of a decreasing series, we have only to subtract the common diffe-
rence from the known term. In this manner, we may find any term
of a series by a continued addition or subtraction of the common
difference. But, as in a long series, this process would be tedious,
it becomes necessary to deduce a formula by which any term, as
the n^\ may be found without going through a tedious addition or
subtraction.
FORMULA FOR THE Nik TERM.
480. The second term of the ascending series, a, a + d, a + 2c7,
&c., is a-\-d; that is, the common difference is added to the first term
one less number of times than the place of the term. The third term
is a + 2^; that is, the first term is increased by the common difference,
taken once less than the place of the term. We discover the same law
of formation in regard to the fourth term, and all succeeding terms.
Hence, calling, the »*•» term 7, we have 7 = a -f (« — 1)^; that is,
the n^* term is equal to the first ternij increased hy the common diffe-
rence, taken once less than the place of the term.
If the series is decreasing, d is negative, and the formula becomes
l=za — in — l)c?. The first formula includes the second ; and the
first, therefore, is only necessary, care being taken to attribute to e2 its
proper sign.
For an ascending series the first term is always the least, and for a
descending series the first term is always the greatest.
The foregoing formula, 7 = a + (n — V)d, may be used to obtain
the last term, for n may represent any term whatever. When the com-
mon difference is equal to the first term, that is, d = a, then, l = a
+ na — a = na. The n^^ term is then equal to the first term into
the number of terms. When the number of terms is unity, that is,
n = 1, then, ^ =: a, as it ought to be.
35*
414 ARITHMETICAL PROGRESSION.
EXAMPLES.
1. The first tcrni of an ascending scries is 2, the common difference
3, required the fifth term. Atu. 14.
2. Required the tenth term of the same series. Ans. 29.
3. Required the last term of an ascending series, of which the firn
term is 10, the common difference 5, and the number of terms 11.
Ans. 60.
4. Required the eleventh term of a descending series, of which the
first term is 60, and the common difference 5. Ans. 10.
5. Required the tenth term of an ascending series, of which the
first term is 10, and the common difference 10. Ans, 100.
6. Required the last term of a decreasing series, of which the first
term is 100, the common difference 10, and the number of terms 10.
Ans. 10,
FORMULA FOR THE FIRST TERM.
431. The formula, I ss a -{-(n — 1)(Z, contains four quantities, and,
of course, if any three are given, the fourth can be determined. Rj
transposition and reduction, we have a = I — (n — l)c^. That is, At
first term is equal to the last, or n**» term, minus the number of ternu,
less one, into the common difference.
The formula is applicable both to an ascending and to a descending
series, by attributing the proper sign to d.
EXAMPLES.
1. The fifth term of an ascending series is 14, and the common
difference 3. Required the first term. Ans, 2.
2. The tenth term of the same series is 29. Required the first term.
Ans. 2.
3. Required the first term of an ascending series, whose last term L^
60, common difference 5, and the number of terms 11. Ans, 10.
4. Required the first term of a descending series, of which the last
term is 10, the common difference 5, and the number of terms 11.
Ans. 60.
ARITHMETICAL PROGRESBION. 415
FORMULA FOR THE COMMON DIFFERENCE.
432. From the fbrmula, ? = a + (» — 1)^, we get <£ s= -^ — -, that
n — 1
is, iJie common difference is equal to (he last term, mintis Hie first term,
divided by the number of terms less one.
For a descendiDg Beries, d is negative, and, by multiplying both
ft ^— 7
members by minus unity, we get d =« =, that is, iJie common dif-
/erence of a descending series is equal to the first term, minus the last
term, divided by the number of terms less one When I =s a, d will be
zero, as it ought to be. When n = 1, <^ = oo . The symbol of absurdity
ought to appear under the hypothesis 71 r= 1 j for, when there is but
one term, there can be no common difference, and therefore the assump-
tion of its existence is absurd.
EXAMPLES.
1. The last term is 50, the first term 10, and the number of terms 3.
Required the common difference. Ans, 20.
2. The last term is 10, the first term 50, and the number of terms 3.
Required the common difference. Ans. 20.
3. The last term is 12, the first term 1, and the number of terms 4.
Required the common difference. Ans. y.
FORMULA FOR THE NUMBER OF TERMS.
433. From I = a + (n — l)d, we get n = 1 H — -z — , that is, the
number of terms is equal to unity, added to the quotient arising from
dividing the difference between the first and last terms by the comman
difference.
The formula is applicable to a decreasing series, by attributing the
appropriate sign to d; when I — a, n will be equal to 1, when <? = 0,
fl. = CX5 .
EXAMPLES.
1. The last term is 50, the first term 10, and the common difference
20. Required the number of terms. Ans. 3.
41C ARITHMETICAL PR0QBE8SI0N.
2. The last term is 12, the first term 1, and the common difference
y . Beqnired the number of terms. Ant, 4.
3. The last term is 50, the first term 10, and the common difference
1. Bequired the number of terms. Aug, 41.
FORMULA FOR THE SUM OF THE TERMS.
434. We might get the sum of the terms by actually performing the
addition ; but, when the series contains many terms, this operation would
be difficult, and we are enabled to deduce a formula which abridges the
work. This formula is deduced from the remarkable property of an
arithmetical progression, that the sum of any two terms at equal distance
from the two extremes is equal to the sum of the two extremes. To
show this, any term, as the m*^, counting from the first term, will be
expressed by a + (m — l)d, and the m** term, counting from the right,
will be expressed by I — (m — 1)^; and the sum of these terms b
plainly a -{- ly a denoting the first term and I the last term.
Calling S the sum of the terms, we have S'=asa+(a+<Q+(a+2«/)
+ (a + 3(f) + C^' + 4(f) .... to Z, and reversing the series, S =
/+(/ — r/) + (/ — 2d) + (I — 3(f) ... to a. Adding these equations
member by member, we get 28 = (a + f ) + (a + f ) + (a + +
&c., up to n terms.
Hence, 2« = (a + f)«, or « = ^ — 5 , that is, the mm of the
id
iermz it equal to the half turn of thefirtt and latt termty into the num-
ber of termt.
When n = 2, « = a + f, as it ought to be. When f = a, « r= na,
the common difference is then zero. When n = 4, 9 a 2 (a + 0^ ^ ^^
ought to be. When n = 0, « = 0, as it ought to be.
The sum of the terms can be determined when a, I and n are known,
or can be found from the data.
EXAMPLES.
1. Find the sum of the natural numbers, 1^ 2, 3, 4, 5, 6, 1, 8, 9.
Ant. 45.
2. The first term of a series is 10, the last term 100, aad the
number of terms 5. Required the sum of the terms.
Ant. 275.
ABITHHETIOAL PBOOKSBBION. 417
8. The fii8t term of an afioending series is 5, the common difference
6^ and the nnmber of terms 5. Bequired the sum of the terms.
Aim, 75.
4. The last term of an ascending series is 60, the nnmber of tenns
4, and the common difference 2. Bequired the sum of the terms.
Am, 228.
I 6. The last term of a descending deries is 60, the number of terms
4^ and the common difference 2. Required the sum of the terms.
jItm. 252.
6. The last term of an ascending series is 50, the first tenn 6, and
the common difference 2. Bequired the sum of the terms.
Aim. 644.
7. Knd «, when a, J, and n^ are known.
A'M. S = }|2a + (n — iy|n.
8. The first term of an ascending series is 1, the common differ-
ence 1| and the number of terms 10. Bequired the sum of the
terms. Aw. 55.
9. How many strokes does the common dock strike in 12 hours ?
A'M. 78.
10. A body falling in vacao will pass over about 16 feet the first
second, 48 the next second, 80 the third second, &c. How far will
it fiill in 20 seconds, and what space will it pass oyer in the last second ?
Am. Entire space, 6400 feet; in last second, 624 feet.
11. A traveller goes 5 miles the first day, 15 the second, 25 the
ihird, &c. Bequired the space that will haye been passed oyer at the
end of the tenth day. Am, 500 miles.
12. Find the sum of the natural numbers, 1, 2, 8, &c., up to n terms.
A'M. } n (n + 1).
13. Find the sum of the odd numbers, 1, 8, 5, 7, &c., up to n terms
A'M, ff.
14. Fbd the sum of the even numbers, 2, 4, 6, 8, 10, &o!, up to n
terms. A'M, n(n •\' 1).
15. Find the sum of the numbers, 6, 12, 18, 24, &c., up V> » terms.
Am. 3n (n + 1).
2b
k
418 ABITHMSTIOAL PBOOBESSIOM.
16. Find the Bum of ihe odd numbers; 1; 2, 3^ &c.; up to 73.
Am. 5329.
17. Find the sum of the even numbers^ 2, 4^ 6; &c.; up to twentj-
fiye tenns. Ans. 650.
18. Find the sum of the numbers, 6, 12, 18, &c., up to the num-
ber 72. Ans. 468.
19. Find «, when I, d and n, are known.
Ans. S = }|2/— (n — l)cf|ii.
TO FIND THE ARITHMETICAL MEAN.
435. The ^rithmetioal mean between several quantities, is the quo-
tient arising from dividing their sum by their number. The arithmeti-
oal mean between two quantities, m and n, is half their sum, that is,
^ (m + m)* We have seen that a geometric mean between two quantitieB
is the square root of their product.
From th9 definition, the arithmetical mean between any number of
tenns in progression must be — = o = K<» + 0> ^^^ *»i ^
n
' ariti^meHeaJ fnean is eguaX to (he half sium of the extremes.
CoroSary.
The first and last terms may be found when the arithmetical
the number of terms, and the common difference, are known.
For, from 8 ^ ^ f we get a = (a + (» — 1) c?), or
A n
a = 2M — a — (n — 1) d. Hence, a = M — ^^^ — - -.
And since 1= a + (n — l)rf, we get / « M + i — ^
The employment of the arithmetical mean frequently fiuulitates
ihe solution of a problem in progression.
AMSHMSIIOAL PBOGBXBBION . 419
IXAMPLK8.
1. The sun of four numbers in aiiUimetical progression is 20^ and
{heir continued product; 884. What are the numbers t
An$. 2, i, 6; 8.
For, M = — = V=6. Andthefiratterm, a = M — ^-=^^^
= 5 — ^; the second term, 5 — id; the third term, 5 + id; the
fourth term, 5 + l^d. Hence, by the conditions of the problem,
(5 — id) (b — id) (5 + id) (5 + id)^ 884, or 625 — ?^ +
Making (r = y, we get, after reduction, y" ^-^ » —,
„ 600_^464 . . ^64 -, , ^ , . . ,
Hence, y = -^ db -^ =+4, or + -^. The last value bemg rejected,
we have cP = + 4. Hence, d = dz2. The positive value will give
a = 5 — |<Z=s2; and the second term, 4; the third, 6; the fourth, 8.
The negative value of d will give the series in reverse order, and we
see here a change of sign foUowed l^ a change of direction.
The same problem might be solred by making 2x the common dif-
ference, and y — 8x, the smaller extreme. Then the 4 terms will be
represented by y — Sx, y — x^y + Xf and y + 8a;. And we have the
two equations, y — 8x+y — as + y + as +y-f8xB=4yss20, and
(y — 3x)(y + 3x)(y— x)(y + a;)-(y»— 9x«)(y«— «0-y* —
lOxV -f ^^ = ^^' Substituting, in this last equation, the value of
y drawn from the first, we get 625 — 250a:' + 9x* ^ 884. From
which, a;« drl. Theny — 8« ^ 5 — 8 = 2, and y — a5« 4, &c.
The student will readily perceive the advantage of repreeentiDg the
first extreme by y — Sx, and the common difference by 2a;.
2. The sum of 5 numbers in arithmetical progression is 15, and their
product 120. What are the numbers ? An$, 1, 2, 8, 4, 5.
420 ARITHMETIOAL PBOGRXSBIOK.
INSERTION OF MEANS BETWEEN THE EXTREMES OF A
PROPORTION.
436. The formnla^ d &= =-, enables us to insert any number of
means^ wHen tbe extremes are known. Let it be required to insert 3
means between 1 and 9. Then a = 1, and l = 9y and, since 3 means
are to be introduced between 1 and 9, n must be 5. Hence, d =:
9 — 1
^ =- = 2. Knowing the common difference, the terms sncoeeding
the first can be readily formed. They are 3, 5, 7 ; and die 5 terms of
the proportion are 1, 8, 5, 7, 9.
EXAMPLES.
1. Required 15 means between 1 and 9.
Ans, d=zi. Series, 1 . 1} . 2 . 2} . 3 . 3} . 4 . 4} . 5 . 6} . 6 . 6} .
7 . 7} . 8 . 8} . 9.
It is plain that an infinite number of means may be inserted between
1 and 9, by making the common difference indefinitely small.
2. Insert 10 means between 1 and 2.
Ans, d = ^j. Series, 1 . ly\. . 1/j . 1^ . ly^ . 1^ . 1/j . 1^^^ . 1^.
lyV . lif . 2.
GENERAL EXAMPLES.
1. Find 4 numbers in arithmetical progression, whose sum is 14,
and the sum of whose squares is 54. An$, 2, 8, 4, 5.
Let y — 3x ss lesser extreme, and let 2x ss common difference.
Then the other terms are y — Xfi/ + x and y + Bx, Hence, by the
conditions, y — 3x + y — a5 + y+ a; + y+ 3aj:«4y = 14, orysBs
3i. And (y — 8x)« + (y — «)«+ (y + x)« + (y + 8x)« « 4y« +
20x' ass 54, and eliminating y", we have 49 + 20x' &» 54, or ^c" ss }.
Then, a; ss d: }. From which the above series.
2. Find 5 numbers in arithmetical progression, whose sum is 40, and
the sum of whose squares is 410. Ans. 2, 5, 8, 11, 14.
Let y — 4x SB lesser extreme, and 2x « common difference. Theo
the terms will be represented by y — 4a:, y — 2x, y, y + 2ic, y -f- 4x.
AEITHMBTIOAL PBOGEISBION. 421
8. Find 4 numbers in arithmetical progression, whereof the product
of the extremes is 700, and the product of the means 750.
Am. 20; 25, 30, 35.
4. The sum of 3 numbers in arithmetioal progression is 18, and the
sum of their squares 116. What are the numbers ? Am, 4, 6, 8.
Let y represent the mean, and x the common difference, then will
^ — X, y viAy'\-Xy represent the three terms of the progression.
5. A and B are separated by a distance of 46 miles. A travels
towards B at the rate of 1 mile the first hour, 2 miles the second hour,
8 miles the third hour, &c. B trayels towards A at the rate of 10
miles the first hour, 8 miles the second hour, 6 miles the third hour,
&c. When will they come together ?
Am, At the end of the fifth hour, and also at the end of the
eighteenth hour.
The second value needs explanation. The formula Z = a -— (n — \)d,
which gives the distance travelled by B during the n*^ hour, shows that
B does not travel at all the sixth hour. And since, when n ^ 6, Z is
native, we conclude that B turns back, and travels in the contrary
direction after the sixth hour.
16inilM
^ A P B M_
At the end of the fifth hour the two travellers were together at P,
15 mUes from A. In 13 more hours the traveller A will be 156 miles
£rom P, at a point M. The traveller B rests during the sixth hour, and
at the beginning of the seventh hour turns back in pursuit of A, and
at the end of the eighteenth hour from the time of his first starting
£rom B, he will also be at M, 156 miles from B. These results are
easily deduced &om the formulas.
When the travellers come together at M , A will have travelled 17
miles, and B 201 miles.
The problem explains most satisfactorily the meaning of a negative
solution, and shows that, when distance is the thing to be determined,
the negative sign always implies a change of direction.
6. A fugitive from justice has one hour the start of the officers of
the law, and travels uniformly at the rate of 7{ miles per hour. The
officers travel 5 miles the first hour, 6 the second, 7 the third, &c. How
36
422 OEOMBTBIOAL f BOaBEBBIOlY.
long will it be from the time of the fugitire's escape until his airest
again 1 Ans. 9 hours, » = 8w
7. The mean of seven tenns in arithmetical. progression is 6, and the
product of the extremes 27. What is the common difference, and
what the series ? Ans. d sa 1. Series, 3, 4, 5, 6, 7, 8, 9.
8. A liquor-seller makes 24 gallons of a mixture of water, brandy,
and rum. The number of gallons of the three fluids constitute a pro-
gression, of which the number of gallons of water is the lea^ extreme,
and that of rum the greatest extreme. The sum of the gallons of water
and brandy is equal to the number of gallons of rain. Seqoixed tibA
quantity of each. Aia, 4, 8, and 12«
9. A man sold a horse upon condition that he should receive 1 cent
for the first nail in his shoes, 11 cents for the second nail, 21 cents for
the third nail, &c., for the 32 nails in his shoes. How much did he
get for him ? Ans. 949 and 92 cents.
10. One hundred stones are placed on the same straight line, 4 yards
apart. How far will a person have to walk, who puts them one by one
in a basket placed on the same straight line, 4 yards from the first stone,
and 8 yards from the second stone : the person being supposed to start
from the basket f Ans, 22 miles, and 1680 yards.
11. Same problem as last, except that the basket is placed at the
first stone. Ans. 22 miles, andi680 yards.
12. Same problem as 10th, except that the carrier is at the other end
of the line, at the 100th stone. Bequired the last term, the ccmunon
difference, the first term and the entire distance ?
Ans. S Si 22 miles, 1280 yards, Z = 8, a » 792.
The progression begins with the second term.
GEOMETRICAL PROGRESSION.
437. A Geomstrical Progression is a series of terms, any one of
which is formed from that which immediately precedes, by multiplying
by a constant quantity. When the constant multiplier is greater dian
OEOMXTBIOAL PBOQEXBIIOir* 423
unity, the series is an inoreasing progression. Thus 1, 2, 4, 8, 16, 82,
64, Ac, constitute an increasing progression, the constant multiplier or
ratio of the progr€$nan being 2. And 82, 16, 8, 4, 2, 1, is a decrea««
ing progression, with the ratio }. The ntio of a decreasing progression
is always unity diyided by an entile quantity. It is evident that aa
increasing progression taken in reverse order will be a deereasiBg pro*
gression, and that the ratio of the latter progression will be t^e recipro*
cal of the former. The converse of this is also plainly tme. Thus h,
-, -g, -^, Ac., is a decreasing progression ^rhose ratio is -, a, being
supposed an entire quantity, and --jf -^ -, and h, constitnte an in-
creasing progression, whose ratio is a.
It is plain that a Geometrical Progression differs from a Geometrical
Proportion only in its number of terms.
FORMULA FOR THE Mh TERM.
488. Let the first term be a, and r the ratio of the progression : then
ar will be the second term, af* the third, at* the fourth, and so on.
That is, each term is equal to the first term, multiplied by the ratio
raised to a power denoted by the number of terms, which precede the
required term. Hence, if I represent the n*^ term, we must have / ^
ar*~\ that u, the n*^ term is equal to the first term, multvplied hy the
ratio raised to the (n — 1) potoer. I may represent the last term, in
which case n — 1 will be the number of terms less one. The formula,
I BBS af^\ trill enable us to determine any term, when the number of
preceding terms, the ratio and the first term are known. When n an 1,
we have I ssa a ; when n sa 2, I ^as ar, &c. When a bb o, / also = o;
when r « 0, /also « o; when a = r*, / =** r".
1. Find the last term of the progression, 1, 2, 4, &c., carried on to
11 terms inclusive. Ans. I =s 1024.
2. Find the n*^ term of the progression, 8, 9, 27, &c. Ans. 3*.
8. Find the 21st term of the progression, 2, 8, 82, 128, ftc.
Am. 219902.8255552.
4. Find the 6ih term of the series, 64, 16^ 4, &c. Ans. -j^.
424 QEOMETRIOAL PEOaBESSION.
The n^ tenn being made up of a product and a power, ib muck
greater in geometrical than in arithmetical progression, when r is ^ 1,
and much less in the former than in the latter, when r<^l. Thus, a
corresponding example to 3, in arithmetical progression, with the first
term 2 and common difference 4, would give 82 for the 21st term;
and the answer to a corresponding problem to 4, in arithmetical pro-
gression, would be 44.
FORMULA FOR THE RATIO OF THE PROGRESSION.
»— 1/7"
439. The equation, I = a7^\ gives r = k/ — , that is, (he ratio u
e^[ual to the (n — 1)** root of the quotient arising from dividing the n**
term hy the 1st term.
This formula can be used when n, l, and a, are known. When ? = a,
I
we have r = 1 : when n = 1, r = oo : when ^n = 2, r = — .
a
EXAMPLES.
1. The 3d term of a geometrical series is 266, and the first term 4.
What is the ratio? Am. r as 8.
2. The fourth term of a geometrical series is 2401, and the first
term 7. Required the ratio. Ans. r = 7.
8. The first term of a geometrical series is 50, and the fourth term
6250. Required the ratio. Ans, f = 5.
4. The first term of a geometrical series is 6250, and the fooith
term 50. Required the ratio. Ans. r = ^.
5. The first term of a geometrical series is a, and the (m -f 1)*^
term of. Required the ratio. Ans. ly oT
CorcUary.
440. The formula, r = k/ — , enables us to insert any number of
means between two extremes. If we wish to introduce m means be-
tween a and /, then the total number of terms will be in + 2. Hence,
OBOMETBICAL PROGREBBION. 425
w+1/7"
fi — 1=*» + 2 — l=m-t-l, and r = \j — . Thos^ let it be
reqxiiied to insert 8 means between 2 and 82 ; then^ r = (/^= V^
= 2. And tbe series is 2, 4^ 8, 16, 82.
EXAMPLES.
1. Find three means between -f^ and 1728, and tbe ratio of the pro-
gression. Afu. 1 . 12 . 144, and r = 12.
2. Find two means between \ and 576000.
Am. 40 and 4800.
FORMULA FOR THE SUM OF THE TERMS.
441. If we multiply a series of terms in Geometrical Progression
by the ratio, a new series will be produced, whose first term will be the
same as the second of the old series, and whose other terms will all
be the same as the corresponding terms of the original series, except
its last term. It is plain, then, that if the old series be subtracted
from the new, all the terms will be cancelled except the first of the old
series and the last of the new.
Take, for example, the series, 8, 9, 27, 81, 248.
And multiply each term by the ratio, 9, 27, 81, 248, 729.
Subtracting the first from the 8, 729.
second series, we have left but two terms.
Let S represent the sum of n terms in geometrical progression.
Then, ^ =^ a '\' ar '\- at* -{• ar^ ar^''\
Multiply by r, and we have ^r =zar •{• ai^ -{- ai^ ar»~* -f ar^,
_ . -. Of (t* — 1^
Hence, Sr — S = ar^ — a, and S = — ^ =— ^.
T — - 1
That is, the turn of the terms it equal to the first term into the diffe-
rence between unity and the ratio raised to a power denoted by the
number o/ terms, and this product divided by the ratio, less 1. This
formula can be used when the first term, the ratio, and number of
terms are known, or can be determined. Since 7 = ar"^', then,
Ir s=r ar^, and the formula may be written S = ^ ; that iS; the
36*
426 OBOMETRICAL PEOOB£B8IOK.
$wn o/ihe terms is equal to the last, or n*", term into the ratw, mtntis
Aejirst term, and this difference divided hy the ratio, less 1.
When n = 0; the seoond members of iJie equations in S and Sr will
cancel each other^ and give S.b* 0; bat. when r »■ 1, the eqnatign in
S becomes S = na, whilst the fonnula^ S = ^ \ becomes S = §.
This is generally the symbol of indetCTmination, bnt^ in the present in-
stance, indicates a. vanishing fraction. The common factor to the two
terms of the fraction is plainly r *- 1, because r — 1 is an exact divisoar
of the nomerator. Performing the division, we have S = — ^ — ~ ^
= a (r»"* + r*-* + r»~* + r + 1). Now, making r = 1,
we have S = na
442. When the progression is decreasing, Sr is less than S ; henoe,
to find S, we must take Sr from S. Then, S — Sr = a — ar^^ or
« a (\ — r») a 1 — Ir ^,_ ^ . ^v - .,
S = — ^ ^ = — = , that IS, the mm of the terms of a de-
X ^— r 1 — "■ r
creasing jfrogression is equal to the first term into unity, mums the hut,
or n*^, term into the ratio, and this difference, divided hy one, misms
the ratio.
EXAMPLES.
1. The first term is 2, th^ ratio 2, and the number of terms 9.
What is the sum of the terms ? Ans. 1022.
2. The first term is 512, the last term 2. Beqnired the soia of the
9 first terms of the progression. Ans, 1022.
8. Bequired the sum of the series, 1 +^ -f-x^ + o:^
a^ — 1
Ans. =~.
« — 1
4. Bequired the sum of the series, a^^ + a^ + .... a + 1,
1 a-
Ans. -= ' .
1 — X
5. Bequired the sum of the series, a;""* + o*^ + x^-y . . . ,
^ ^ Ans. ^ , or =^.
y — X X — y
6. Bequired the sum of the series, a;' + ox + a'.
Am. , or ^.
a — a; a; — a
GBOMXTBIOAL PB0QBBS8I0V. 427
(f» 1\
443. The two equations^ I = af^\ and « = a ^ ^ , contain
fiye quantities^ any two of which can be detennined when the other
diree are known. Thus, wh«K a, I and n, are known, we obtain^ by
combining these equations, and eliminating r, « = - =.
In like manner, by eliminating Z, we get a » • J^i ^ ^^^i ^7 ^*
(y^ ,^>\,
minating a, we find I b ^-— ^ .
AN INFINITE DECREASINQ PROGBESSION.
444. The formula for the sum of the terms of a decreasing progres-
fflon takes a remarkable form when the series ia infinite.
Por, S = -= may be written, 8 = r ■= .
Now, since r is a firaction whose numerator is unity, like the firaction
\j it is plain that, when n bs op , r^ will be zero. Hence, .. ■ ■ will
be- zero, and that part of the formula may be neglected when the num-
ber of the terms is infinite, and we may write 8 » / that is, the
turn of the ternu of an infinite decreasing progression is eqwd to the
Jirst term divided by unity y minus the ratio of progressions.
This formula will give the Ztmt^ of tibe series, that is, a value which
the sum of the terms cannot exceed. Thus, the decimal '33333, ftc.,
may approach infinitely near to •), but can never exceed it
BXAMPLE8.
1. Knd the limit of the value of the decimal, -666666, &c.
Ans. I
For a = ^, and r = ^] hence, S = ^ ^ =4=1-
^ TIT
2. Find the limit of the value of the decimal, *1111111, &c.
Ans, ^. •*
8. Ym^ the limit of the value of the decimal, '66565, &e.
428 GEOHETRIOAL PBOGBESSION.
4. Find Uie limit of the yalue of -99999, &o. Aiu. 1.
6. Find the limit of the value of *88888| -&o. An». {.
6. Find the limit of the value of theseries, ^"^^ + 19+ sV + '^^
An$. l+i-
7. Find the limit of the value of i + r^^ + ^^ &c. Ans, }.
8. Find the limit of the value of the series, 20 + 4 + 1 + 3^+^
^fw. 25.
9. Find the limit of the value of | + ^^ + jis + &o.
Ans. 1.
10. How much will the sum of the series, | + jf^, &c., differ from
unity, when four terms only are taken f An$. By ^^
Use the expression, ^ , which can only be neglected wlien n sb oo .
11. Find the sum of the series, 42 + 6 + f + 7^+ if^i &c.
Am. 49.
12. Find the limit of the value of 6 + f + :,<^ + ,|, + &o.
Ang» 7.
ft
18. By how much will the sum of the series, 6 + 4 + A + iff +
&o., differ from 7, when three terms only are taken.
Am. Bj -J^.
GENERAL EZABCPLES.
1. The first term of a geometric progression is 4, the last tenn 32,
and the number of terms 4. Bequired the sum of the terms.
Ang. 60.
2. There are three quantities m geometric progression, their sum is
a, and the sum of their squares, V, What are the quantities ?
a« — 6« a" + ft" v^lOa»6» — 36* — 3a*
^^'•y 2^^^ 4^+ iS ''-
a« -f y ^lOa'6' — 3&* — 3a*
4a 4a
For, let Xy y and z, denote the three quantities. Then, x 4- y + '
=afj^=i xZf and a^ + y" + «■ = ft". Transposing y to the second
member, and squaring the first equation, we get cc* + 2xz + 2'=a' —
2<^y+^) which, combined with the second, gives a:* + ^ + 2" = a* —
OXOUETBIOAL PBOGBESSION. 429
2ay. Subtract tbe third equation from tbiS; and soIto witb reference
to y; we will find then^^ « -— . The third equation gives a' + «"
s= V — y*. Subtract from this the equation, 2xz = 2^". Then, a^ —
2xM + s^zssh^ — 3^, and, taking the square root of both members, we
find, X — 2f = v'^* T— 3^ (M) ; but, from the first equation, as + « =
a — y (N). Adding and substracting (M) from (N), we get x =
2 =""i^+ 4a , and.=
a —y _ ^y _ 3y» _ a' + fe' _ v^lOa'6' — 3fe* — 3a*
2 "'4a 4a '
The problem might haye been solved by the ordinary method of eli-
mination, but it would have led to an equation of the fourth degree.
8. Two travellers were separated by a distance of 36 miles. The
one in advance travelled 1 mile the first day, 5 miles the second day,
9 the third day, and so on. The one in the rear travelled 1 mile the
first day, and increasing his rate in a geometrical ratio, travelled 64
miles on the seventh day, when he overtook the advanced traveller.
Sequired the uniform rate of increase of the second traveller's duly
distance. Aiu, r = 2 miles.
4. Find v when 9, a and I are known. Am. r b ^
Problem 3 is but a particular case of problem 4.
5. There are four quantities in geometric progression, the sum of the
first two is a, and of the last two, b. What is the ratio and what are
the numbers ?
/T ^ a ^V a b
Ant. r SK V / — . Numbers, =., =, =,
and
w^
6. Tbere are four numbers in geometric progression, the sum of the
first two is 30, and of the last two, 750. What is the ratio, and what
are the numbers ? Am. f ss 5. Numbers, 5, 25, 125 and 625.
480 aXOMBTBICAL PBOOBSSSIOlff.
7. There are three quantities in geomeirio progression, the first is a,
and the second h. Bequired the third. . h^
a'
8. There are three nnmbers in geometric progression^ the &Bt is 4|
and the second 20. Bequired the third. An$. 100.
9. Find a, when ^ r and S are known.
Am. a«sS — (8 — l}r.
10. The ratio of a geometric progression is 5, the sum of the terms
780, and the last term 626. Required the first term. Ans. 5.
11. Find Z; when a, r and S are known.
12. A rich, but charitable man, gave $20 to the American Tnei
Society, twice as much to the Board of Foreign Missions, four times a
much to the Board of Domestic Missions, and so on in the same niia
His last contribution was to bis own church, and his entire charitj
amounted to 11020. How much did he give his church ?
Ans. $520.
18. A man sold a horse upon the following terms : He was to recem
one cent for the first nail in the horse's shoes, two cents for the seoond
nail, four eents for the third, and so on fot the 32 nails in the hone's
shoes. Bequired the price of the horse.
Am. 4,294,967,295 cents.
14. A gentleman made a donation to a charitable institution, upoo
the following conditions : Five hundred dollars were to be receiTcd the
first year, $250 the second year, $125 the third year, and so on forerer.
Bequired the amount of the donation. Ans. $1000.
15. A gentleman wishes to bequeath $3000 to a Beneyolent Assoda-
tion in the following manner : Two thousand dollars to be given the
first year to meet the present wants of the association, and ihe donation
every year after, forever, to be in a decreasing ratio. What must the
ratio be 7 Ant, i.
16. A gentieman invested $1000 for the benefit of a charitaMe
institution, so that one half less should be drawn from it each year than
the preceding year, forever. How much must be drawn the first year?
Ans. $500.
INSQII1.LITIES. 481
INEQUALITIES.
445. An inequality is an expression to signify that two quantities are
Tinequal to each other. Thus, A ^ B, is an inequality indicating that
A is greater than B. For equalities, it matters not on which side of the
sign of equality the two quantities are written. Thus, A = B may be
written B 8s A. But, for inequalities, there can be plainly no trans-
position of the two quantities without a corresponding inyersion of the
sign. In the expression, A ^ B, if A be changed to the second mem-
ber, and B to the first, there must be an inversion of the sign. For, let
m be the quantity which, added to B, makes it equal to A, then, A «
B -f m, or B + m « A. Then B <^ A. So, B and A have changed
places, with a corresponding inversion of sign.
Inequalities are used to determine the limits between which quantities
are found, and are of frequent application in the higher mathematics.
As a simple iUustradon, suppose we have the two inequalities, x ^ 5,
and X <[ 10. We see that x must be some number between 5 and 10,
and these numbers are the limitt to its values. When the enunciation
of the problem restricts the solution to positive quantities, there are but
two limits, the superior positive limit, and the inferior positive limit.
So, if the solution be restricted to negative quantities, there are but two
limits, the superior negative limit, and the inferior negative limit. In
general, however, there are four fimits, two superior and two inferior.
Two inequalities are said to subsist in the same sense, when the
greater quantity lies on the same side of the sign in both of them; and
they subsist in a contrary sense, when the greater quantity in one ine-
quality lies on a different side of the sign firom that occupied by the
greater quantity in the other inequality.
Thus, 4^2, and 5 ]^ 3, subsist in the same sense; so also, 2 <^ 4,
and 3 <^ 5. But 4 ]> 2, and 3 <[ 5, subsist in a contraiy sense.
When two quantities are negative, that one is the least, algebraically,
which contains the greatest number of units. GThus, — 15 <^ — 10,
— 5> — 7.
446. There are ten important principles, belonging to the subject of
inequalities, which require to be demonstrated.
1. If the same quantity be added to, or subtracted from, the two mem-
bers of an inequality, the resulting inequality will subsist in the same
sense.
482 INEQUALITIES.
For, let A ^ B, and m the differenoe between A and B, then, A =
3 + m. Now, add or snbtraot the same quantity, c, from both members,
and there results Ad=C8sB + md=c. Hence, of couise, A±e^
B db c, and the inequality subsists in the same sense.
This principle enables us to transpose terms from one member of the
inequality to the other. Take, x + 2 ^ a ; add — 2 to both members,
and there results x^a — 2. So, transposition is effected in inequaliti^,
as in equations, by a change of sign.
2. If the two members of an inequality are multiplied by a positiTe
quantity, the resulting inequality will subsist in the same sense.
For, let A ^ B, and AsssB -{- m. Multiply both members by r,
and we get Ac b Be -f mc. Hence, Ac ^ Be. This principle serves
to free the fractions, if any, of their denominators.
Take -^ T^^- Clearing of fractions, there results 2x — a >
3. If both members of an inequality are divided by a pomtiTe quan-
tity, the resulting inequality will subsist in the same sense.
For, let A ^ B, and m the difference between A and B. Then A =
B + m. Dividing both members by c, there results, — s= (- — ,
c c c
Hence, — j> — .
The last principle serves to clear the unknown quantity of its coeff-
a -h4y
dent when positive. Take, 2x ^ a + 45*, then, x ^ — - — .
The three foregoing principles serve to solve inequalities, when tiie
coefficient of the unknown quantity can be made positive.
4. If two inequalities, subsisting in the same sense, are added
together, member by member, the resulting inequality will sabsst in
the same sense.
For, let A > B, and C> D. Let m be the difference between A
and B, and n the difference between G and D. Then, A s= B + m,
and C » B + n. Adding the two equations, member by member,
there results A + C = B + D + m + n. Hence, A -f C >► B + D.
5. If an inequality be multiplied by a negative quantity, the result-
ing inequality will subsist in a contrary sense.
For, let A > B, or A s= B + w. Multiply both members by — c,
and there results —Ac saa — Be — m, or m — Ac =» — Be Hence,
INEQUALITIBS. 433
Ac being nnmerioaU j greater ihui Be, we have, algebndoalljy — Ac <C
— Be.
The foregoing principles enable ns to eliminate between inequaUties
88 between equalities^ when we have two inequalities involying two un-
known quantities whose signs are known^ and that of one of them in
one inequality oontrary to that of the same unknown quantity in the
other inequality.
Take a? + y > 12,
aj— y > 8.
Adding member by member we get x }> 10, and this substituted in
the first equation, gives y^ 2. Let a; s 10 + m. Then, from the
2d equation y ^2 -{- m. The value of y is then fixed between the
limits of 2 and 2 -f m.
6. If two inequalities, subsisting in the same sense, are subtracted
member by member, the resulting inequality may subsist in the same or
a oontrary sense.
For, let A > B, and C > D. Then, A = B + m, and C = D + n.
Subtracting member by member, there results A — = 6 + m —
(D + n)- or, A — C = B — D + m — n.
Now, if *» > n, A — C will be equal to B — D increased by a posi-
tive quantity, and, of course, will be greater than it. But, if n ]> m,
A — C will not be equal to B — D, until it has been diminished by the
difference between n and m.
In that case, A — C < B — D.
Take, 8 > 4, m = 4,
6 > 8, n » 8.
8— 6>1.
The resulting equaliiy subsists in the same sense, since m^ n.
But, take 8 > 6, m » 2,
6 > 1, » = 6.
8 — 6 < 5.
The resulting inequality subsists in a contrary sense, since n^m.
8. If both members of an inequality are essentially positiye, they
may be squared without altering the sense of the inequality.
For, let A > B, or A = B + f», then A« = B« + 2Bi» + m«.
Hence, A* > B*.
37 2o
484 INEQUALITIZS.
9. If the two membeiB of an inequalily hare oontzaxy agnB| the re-
sulting inequality, after squaring, may subsist in the same or a oontniy
sense.
For, let A > — B, or A « — B + f». Then, A* = B* — 2Bm +
m\ If m" > 2Bm, then, A«> B«; for, A* is equal to B», inoreaBed
by the difference between m' and 26m. But^ if m' <[ 2Bm, then, A*
<[ B* by the difference between 2Bm and m".
Take, 3 > — 2j then, m = 5, m« = 25, and 2Bm = 20.
Squaring, we haye, 9 > 4, since w" > 2Bm.
But take, 3 > — 6; then m « 9, m» = 81, and 2Bm = 108.
Squaring, we get, 9 < 36, an inverted sign, since m' < 2Bm.
When, therefore, the sign of either member is unknown, it is not
permitted to square the inequality.
10. When the two members of an inequality are divided by a nega-
tive quantity, the resulting inequality will subsist in a contraiy sense.
For, let A ^ B, or A SB B + m. Dividing both members by — Cj
there results — — = — . Since ^ — is numeiicallv crreafter
c c c c '' ^
than — — , it must be algebraically less,
c
For equations of high degrees there are generally four limits : a supe-
rior positive and a superior negative limit, and an inferior positive and
an inferior negative limit. But the conditions of the problem may re-
strict the solutions to two limits, and even to one limit.
A single equation of the first degree can have but one limit, this may
be superior or inferior, positive or negative. Limits are determined by
means of inequalities.
SXAMFLEB.
1. The value of x in an equation is such that, twice the value in-
creased by unity cannot be less than 7. What b the lower limit of
the value ? We have, from the conditions, 2a5 + 1 > 7. Hence, as > 8.
The inferior positive limit is then 3, and any number above 8 nuy be
the value required.
2. The value of x in a simple equation of the first degree is known
to be such, that twice the value, plus unity, is greater than 7, and tliat
INEQUALITIES. 485
tiaee times the yalne^ diminished by 4^ is less tlian 11. Between wliat
limits does the yalue lie ?
Ans. Superior positive limit, 5 ; inferior positiye limit; 8.
If the value be known to be a whole number^ 4 mnst be that number.
3. A person desirous of giving some cents to a certain number of
beggars, found on examination that, to give them three cents apiece,
would require more than double of what he had about him, and that, to
^ve them 2 cents apiece, would require more than the difference be-
tween all he had and 35 cents. How many beggars were there, and
how many cents had the person ?
From the oonditiims, Sx ^ 2y,
and, 2x^35 — y.
Multiplying the second inequality by 2, and adding member by men
ber, there results, 7x ]> 70, or a? ^ 10. This value substituted in
ihe first inequality, gives y <C ^^ + I* ; ^ representing the excess of
X over 10. The value of x, substituted in the second inequality, gives
y > 85 — 20 — 2m; or y > 15 — 2m. Suppose m = }, then,
y < 15f , and y > 14.
4. A person desiring to give some money to 11 beggars, found that,
to give them 8 cents apiece, would require more than twice as much
money as he had ibout him, and that, to give them 2 cents apiece,
would require more than the difference between 87 cents and the num-
1>er of cants about him. Required the number of cente he had.
Ans. X <^ 16}, and as ^ 15 ; hence, x « 16.
The nature of the problem makes the solution exact in this case, but
it is very seldom so.
5. Knd the negative limits of x in the inequalities,
aj + 8< — 5, ^
a — 8> — 7.
An$. x<^ — 8, and ar ^ — 4.
6. Find the limit of the value of x in the inequality, ax+b + e
•^ An$. Infenor limit, —-^r-
486 QEMERAL THEORY OF EQUATIONS.
GENERAL THEORY OP EQUATIONS.
447. The general theory of equattons baa for its object, tbe inyesti-
gation of properties common to equations of eveiy degree and of evexj
form.
We will confine ourselves mainly to tbe examination of equations in-
Yolying but one unknown quantity.
Tbe most general form of an equation of tbe m^^ degree witb one
unknown quantity, is a;" + Pa"*"' + Qr""^ + Rx"^ + Tz
+ U = 0.
Tbe coefficient of tbe first term is plus unity, and tbe otber coeffi-
cients, positive or negative, entire or fractional, rational or irrational.
A value bas been defined to be tbat wbicb, substituted for tbe un-
known quantity, will make tbe two members equal to eacb otber. Since,
in tbe general equation of tbe m}^ degree, tbe second member is zero, a
value substituted for x, must reduce tbe first member to zero akoi
Hence, in our discussion, we may define a value to be tbat wbicb, sub-
stituted for tbe unknown quantity in tbe equation, will reduce tbe first
member to zero.
GENERAL PROPERTIES OP EQUATIONa
First Property,
448. If any quantity, a, be a value of x in tbe equation, a^ -^ Px^'
+ . . . Tx + U = 0, tbe first member of tbis equation will be exactfy
divisible by x — a.
For, let Q' be tbe quotient resulting from tbe division of tbe first
member by a; — a, and let R' be tbe remainder, if any, after division.
We sball tben bave tbe identical equation, x" -f Px*~* + Qx"'-"-!- ....
+ Tx 4- U = (y (x — a) -f R'. But, for x = a, tbe first member is,
by bypotbesis, equal to zero, and tbe second member reduces to R^
Hence, we bave = -f R', and, iberefore, R' s 0, and tbe division
is exact
Second Property,
449. If tbe first member of an equation, of tbe form of x" -f Pz*"*
+ Qx"~" -f- . . . . Tx + U = 0, be divisible by x — a, tben a will be
a value in tbis equation.
GXNEBAIi THBOBT 01* EQUATIONS. 437
For, oaUing Q' the quotient, we will have a" + Px*-' +Qx'"*^ +
. . . . Tx + U = Q(x — a). Hence, then, Q'(x — a) s 0. But, when
the product of two factors is equal to zero, the equation can be satisfied
by placing either factor equal to zero. We haye then a right to place
X — a Bs 0, from which there results x ^ a. Therefore, a is a value
in the equation, Q(x, — a) = 0, and, consequentlj, in the given
equation.
CoroUary,
1st. It follows that, in order to ascertain whether any polynomial is
divisible by x — a, we have only to substitute a for x, wherever x
occurs, and see whether the polynomial reduces to zero.
2d. Hence, also, if a polynomial is divisible by x — a, a will be a
value in the equation formed by placing the polynomial equal to zero.
3d. We may also diminish the degree of an equation, when we know
one or more of its values, by dividing its first member successively by
the binomial factors corresponding to these values. The division by
each binomial factor will reduce the degree of the equation by unity.
4th. Numerical equations can frequently be solved by means of the
first two properties. Literal equations can also be solved in the same
way, but more rarely : we have only to ascertain what number or quan-
tity will satisfy the equation ; then, by dividing out the binomial factor
corresponding to this value, we will have a new equation of a degree
lower by unity. A second value may be found from this equation, and
the factor corresponding to it divided out. Thus, we may continue the
process until the given equation is reduced to one of the second degree,
which can be solved by known rules.
EXAMPLES.
1. Solve the equation, x* — 6x' + llx — 6 = 0.
We see that x = 1 will satisfy this equation. Hence, x — 1 is a
divisor. Dividing by x — 1, we get a quotient, x" — 5x + 6 = 0, by
solving which we get x = 2 and 3. Hence, the three values are 1,
2, and 3.
2. Solve the equation, x* — 5x* + 4 = 0.
We find + 1 to be a value, and, dividing by x — 1, we get a new
equation, x" + x* — 4x — 4 =, which is satisfied for x = — 1.
Dividing by x + 1, we get a new equation, s? — 4 = 0, which gives
37*
488 OXNIBAL THSORT OV SQUATI0N8.
the two -nines, x=: + 2, and — 2. The four yalaee are then, + 1,
— 1, + 2, and— 2.
3. Sohe the equation, a^ + 3x* — 5aj» + 4c + 12 — 15x« = 0.
Aru. + 1, — 1, + 2, — 2, and — 3.
4. Solve the equation, aj* — o'x* — 6V + a'ft' = 0.
Ans. + a, — a, +6, — b.
6. Solve the equation, cc* — aV — 6'x' + aV -f JV — aj* + aVx —
a'J* = 0. Ans, + a, — a, -\- h, — h, and + 1.
6. Solve the equation, a^ + a^ — 4x — 4 = 0.
Afu. + 2, —2, —1.
Third Property.
449. Every equation, with one unknown quantity, has as many
values for this unknown quantity as is denoted by the degree of the
equation, and has no more.
Let us assume the equation, cc" -f Px*^' -f Qx"^ + . . . Tx + U
= 0, which is of the m^^ degree. It is to be shown that it has «i
values and no more. We will also assume that eveiy equation has at
least one value. Let a be one value, then x — a is a divisor. Let
a:^' + Faf*-* + Q^aJ*^ + Tx + U', represent the quotient of
the division of the first member by x — a. Then the given equation
will assume the form, (x — a) (x"~" + Px""-* + Qfoc^^ -f &c.) =
. (A), in which, the coefficients, P, Q', are different from the
original coefficients, P, Q, &o. Now, since, in equation (A), we have the
product of two factors equal to sero, we have a right to place either
equal to aero, let us place x""* + Fx"~" + Q^x"*"" -{- &c. = (B).
Equation (B) will also have one value ; suppose it negative and equal to
— b. Then, by the first property, x — ( — 6) = xH-6 will be a
divisor. The first member of (B) can then be decompoeied into two
factors, one of which will be x + ^y &ud the other the quotient arising
from the division of the first member hj x + b. Hence, (B) becomes
(x + 6) (x-^ + F'x— » + Qt'vT^ + &o.) = (C). And (A) can be
put under the form of (x — a) (x + b) (x*""" -h F'x"*~* + Q"x*^ +
&c) =0. It is plain that, by putting the second fiictor of (G) equal lo
lero, we will get another value, and, consequently, another divisor. By
GENEBAL THSO&T Of EQUATIONS. 489
oontiniiing this proceas, ibe degree of as in the successiye quotients will
be dimisisiied by unity eaoh time^ and^ afiber m — 1 diTiabns, we will
obtun a quotient of the first degree in x, from wUch^ of course^ one
value of X will be found. We supposed the first value posdtiYe, and
the second negative; let us attribute the double sign^ db , to the remain-
ing values^ Cj d, e, &o.f since they may be either positive or negative.
The first member of the given equation can then be decomposed into m
binomial factors of the first degree in x, and we will have, x^ + Px"^*
+ Qr--"+ +Tx + U = (a: — a) (x + h) (x ::p e) (x zp d)
&o. = 0.
But, since each &ctor corresponds to a value, and since there are m
diviiorSf or fiurtorS; there must be m values.
Scholium,
1st Had we known that the first member of an equation of the m^^
d^ree could be decomposed into m binomial factors of the first degree
with respect to x, we could readily have shown that the equation must
contain m values. For, it can be satisfied in m ways, by placing each
of its m factors equal to zero, and each fiEictor, so placed, will give a
value. Moreover, since the equation can be satisfied only in m ways,
it contains no more than m values.
2d. It does not follow that all the values must be different Any
number of them, even all of them, may be equal. Thus, the equation,
of — 2rc + 1 ss 0, contains two values, ^ach equal to + 1* The equa-
tion, (x — a)^ = 0, contains m values, each equal to + a. The equa-
tion, (x — ay^(x+by=Of contains m values, eaoh equal to + a, and n
values, each equal to — h. And so for other equations.
3d. An equation of the third degree, such as (x — a) (a:-— 6) (x— c)
= 0, will contain three divisors of the first degree, (x — o), (x — 6),
and (x — c); three of the second degree, (x — a) (x — J), (x — a)
(x — c), and (x — h) (x — c); and one of the third, (x — a) (x — 6)
(x — c). These divisors are evidently equal to the number of combi-
nations which can be formed by combining 3 letters in sets of 1 and 1,
2 and 2, 3 and 3. So, likewise, if we take the m factors of an equation
and multiply them two and two, three and three, &c., we shall evidently
obtain as many divisors of the second degree as we can form combina-
tions of m letters, taken two and two ; and as many divisors of the third
degree as there are combinations of m letters taken three and three,
440
GENKBAL THEOKT OV EQUATIONS.
A;o. The giyen equation will then haye m divisors of tiie fint d^;ree|
m(m — 1) mCm — l)(m — 2)
in Xf — - — ^ — diviBoiB of the seoond degree, — ^ — = — ^-^
of the third degree, &o.
Fourth Proper^.
451. All the coefficients, after the first, of an equation of the m^
degree, are fnnctions of the valaes.
Sappose the general equation of the m}^ d^ree, a^ + Px^"' +
Qz"*"^ -f . . . . + Tx + U = 0, contains the m values, db a, d= ft, ±e,
zh d, &c. The equation can then be put under the form of (x zp a)
(x =f= 6) (x q= c) (a i^ (^), &o., to m factors = 0. By actuaJlj per-
forming the indicated multiplication, we will get,
a
b
c
d
&c.
ifoc
if&c.
It 6c
&c.
T^Zf.
dbc
ahd
acd
&c.
bed
&c.
zp abed, &c. S3 0.
The upper row of signs belon^ng to the positive value, and the
lower to the negative.
We, observe the following rektion between the coefficients and the
values:
1. The coefficient of the second term is the sum of all the values of
the unknown quantity, with their signs changed.
2. The coefficient of the third term is the sum of the products of aD
the values, taken two and two, with their respective signs.
3. The coefficient of the fourth term is the sum of the products of all
the values, taken three and three, with their signs changed.
4. The last term is the continued product of all the values of the
unknown quantity, with their respective signs if the degree of the
equation be even, or with their signs changed if it be odd.
We have supposed in the preceding demonstration that all the vahies
were positive or all negative. But the law of formation for the coeffi-
cients is evidently the same when some of the values are positive and
some negative.
aSNEBAL THSOET OV BQUATI0N8. 441
Thna, let the Talaes he + a^ +h, and — e.
X + ahc s 0.
Then^ (x — a) (x — h) (x + c)ssO, ox a? — a
— h
+ c
flc' + oft
— ac
— he
And we see tliat coefficients are formed in accordance with the above
law.
452. The preceding properties show several important things in
relation to the composition of an equation.
1. If the coefficient of the second term of an equation be zerO; and;
oonseqnentlj, the second term be wanting, the sum of the positive
values must be^ual to the sum of the negative values.
2. If the signs of the terms of an equation be all positive, the values
must be all negative. For^ an equation cannot have all its terms posi*
five unless its binomial factors are of the form, (x + a) (x + h) (x + c)^
&Q, ; which factors, placed equal to zero, will give the negative values^
-^ a, — hf — c, &c.
3. If the signs of the terms of an equation be alternately + and — ^
the values of the unkno^ni quantity will be all positive. For, in this
case, the first member of the equation must be made up of the factors^
(x — o) (a; — h) (x — c), &c., corresponding to positive values.
4. Since the last term, irrespective of its sign, is the continued pro-
duct of the values, it follows that, when the last term is zero, one value^
at least, must be zero.
5. It follows, also, from the composition of the last term, that, in
seeking for a value, we need only seek among the divisors of the last
term, since eveiy value must be a divisor of that term.
6. If we kDow one value, the coefficient of the second term will give
the sum of all the rest, and the last term, divided by this value, with or
without its sign changed^ will give the product of all the rest
CoroUary.
' 453. 1st. The last two consequences enable us to solve equations
with facility when all the values except two are known.
Take, as an example, the equation, as' — \a? — jx + ^ = 0.
The values must be sought among the divisors of the last term:
+ 1 is a divisor of the last term, and may, therefore, be a value ; upon
442 OSNXBAL THEORT OT XQTJATI0H8.
fcrial, we find that it will satasfy the equation, and u, thereforBi a Tiloe.
Calling the other two valnes x and y, we will have, — (1 + ^ + ^) =
^ I, the coefficient of the second term. Hence, x + y ss — |. We
xy
have, also (sixth consequence), -7^ =» — i' Combining these two
equations, we get x ( — j — «) = — ^, or oc" + ^x = + |. And
the values of x and y axe found to be + j, and — ^. The equation
has, of course, but three values, since it is an equation of the third
degree, (Art. 450).
2d. When all the values are known except two, we can tell whether
these values are real or ima^nary without actually finding them.
Thus, take the equation, a* — 6a* — 16x + 21 « 0.
The exact divisors of the last term are + 1, + 3, and +7, — 1,
— 8, and — 7 ; the values must be sought among these numbers.
We find, upon trial, that + 1 and + 3 are values. The second tcna
of the equation being wanting, the sum of the other two values must
21
be — 4, and their product =~-^ =b 7. But 4 is the greatest product
1.0
which can be given by two numbers whose' sum is 4. Hence, the
other two values of the equation are imaginary. And^ in fact, by pur-
suing the preceding process, we find them to be — 2 + ^ — 3, and
_2— v'lrj
EXAMPLES.
1. Solve the equation, ad* — 2aa* — 9a*x + IM = 0.
Ans. + 2a + Sa, and -^ 3a.
2. Solve the equation, a? — ^a? + 0? — 4 ■= 0.
Ans. + 4, +^ — 1, and — ^ — 1.
3. Solve the equation, x* — 2x' — 8 = 0.
Ans. +2,-2, + v'— 2, — v^— 2.
454. 3d. If the values are known, the equations which give those
values can be formed in two ways, either by multiplying together the
binomial factors corresponding to those values (Art 450), or by fimn-
ing the coefficients, (Art. 451).
GINE&AL THEORY OF IQITATIOIfS. 448
SXABiPLSS.
1. Form the equation wHose values are + 1, — 1 and — 2.
Am. a^+ 2a? — X — 2:^0.
For the factors^ multiplied together, and placed equal to zerO| give
(x — V) (x + 1) (x + 2) « (A) or x* + 2a? — x — 2 = 0.
We see that the three fitotors of (A), placed separately equal to lero,
will give the yalues, + 1, — 1, and — 2, or the problem may be solved,
by Art. 450.
Coefficientof 2d term = — (+1 — 1 — 2)= + 2
« 3d term = (+1) (-!)+(+!) (-2>f (-1) (-2)=-l
« 4thteim= — (+1)(— 1)(— 2) -.—2.
2. Form, by both methods, the equation whose values are + 4,
+ v' — 1, and — -v/ — 1- -^*w. a? — 4a;* 4- x — ♦ 4 = 0.
3. Form , by b oth m ethods^ the equations whose values are + 2, — 2,
+ ^—2,-^—2. Ans. «* — 2a:« — 8 = 0.
4. Form, by both methods, the equation whose values are — 1, — 2,
— 3, and— 4. An$. a? + 10x» + 35x* + 60x + 24=0.
Fifth Property.
455. Every equation may be transformed into another, in which the
values of the unknown quantity shall be equal to those of the proposed
equation, iDcreased or diminished by a certain quantity.
Let the given equation be
iB" + Par-* + Qx^ + + Tx + U = 0,
and let it be proposed to transform it into another in y, so that y =
X =b a. The principle of transformation is, of course, the same when
y == X + a, or B X — a; we will then confine our discussion to the
case, when y » x — a. From this equation there results, x r= y -f a.
Substituting this value in the proposed equation, it becomes (y + a)"
+ P(y+a)«-» + Q(y +a)"^+ T(y + a)4-U=0.
Developing the different powers of y + a by the binomial formula, and
arranging the development according to the descending powers of yi we
have the transformed equation,
444
y" + ma
+ P
OINXBAL THKOBT OT IQUATIOHS.
, w (w» ^ 1) g*
+ m (m — 1) Pa
+ Q
+ «-
+ Pa
+ Qo
+ Ta
+ U
t— I
= 0.
If the valnes of y in this equation can be fonnd; the oonespondiii^
yalaes of x in the given equation will be eqnal to these values in*
creased by a.
CoroUary,
456. The preceding transformation enables ns to reduce the number
of terms in an equation. For^ since the quantity, a, is entirely arbi-
trary, such a value may be given to it as will reduce the coefficient q€
any term in the transformed equation to zero, and consequently make
the term itself disappear. Suppose we wish to free the transformed
equation of its second term; we have only to place the coefficient of that
term equal to zero, and find the value of the quantity, a, that makes it
zero. Of course, then, this value attributed to a, will cause the dis-
appearance of the second term of the equation. Placing ma 4-^=0,
P P
we get a = •— — . Then, x = v . Hence, for transforming one
m m
equation into another, in which the second term is wanting, the following
BULE.
Substitute for the wnhnown quantity a new unknown quantity y
nected with the quotient arising from dividing the coefficient of tht
iecond term of the given equation, with it9 sign changed by the degree
of the equation.
EXAMPLES.
1. Transform the equation, x* — 4x = 6, into another, in which the
second term shall be wanting. An$. x=^y — ( — |) = y + 2.
The transformed equation will then be (y -}- 2)^ — 4 (y + 2) a= 5, or
y* — 4 = 5. Hence, y" = 9, and y = d: 3 ; and, a; = y + 2 = 5,
or — 1 ; the same result that we would obtain by solving the given
equation.
OXNZBAL THEORT OT EQITATIONB. 445
2. Transfonn the equation^ a^ — Sx* = — 2, into another, in irhioh
the second tenn shall be wanting. Ans. y*— 8^ «s 0.
The transformed equation can be readily solved, one value of y being
zero. Hence, one value of the given equation is 1. The transformed
equation, in this case and in many others, is simpler than the given
equation, and, therefore, more readily solved. The chief object of the
tnnsformation is to simplify the form of the equation.
Scholium.
457. 1. The third term of the transformed equation may be made to
disappear by giving to a such a value as will satisfy the equation,
— ^ o~^ — + 0^ — 1) Pa + Q = 0. And, since this equation
is of the second degree in a, there are two values for a, and, con-
sequently, the third term can be made to disappear in two ways. In
like manner, the fourth term can be made to disappear in three ways,
the fifth term in four ways, &c. ; and the last, or (m + 1)^^ term, in m
ways. By recurrence to the derivations of these coefficients from the
values, we see that the above results are true. The last term, for in-
stance, being made up of the product of the m values, can be made to
disappear by placing either of the m values equal to zero.
458. 2. It may happen that the value of a, which makes the second
term disappear, will also cause the disappearance of the third or some
other term at the same time. Let us examine under what circumstan-
ces the second and third terms will disappear together. By placing the
coefficient of the second term. Pa + m, equal to zero, we get a =
, and, by placing that of the third term, — i-^= — ^-^ f- (m — 1)
P /P* 20
Pa -t- Q, also equal to zero, we get a = =fc \/ — ..
m \ m m[m — 1^
Now, it is plain that the values of a in this equation will be identical
P
with the last, that is, both , when the radical disappears. Placing
2mQ
the radical equal to zero, we get P = -^. Whenever, then, the
square of the coefficient of the second term is equal to twice the quo-
tient arising from dividing the product of the degree of the equation
38
44S GBNXBAL THEOBT OV SQUATIOVS.
into the coefficient of tHe third term by the degree of the equation,
less one^ the second and third terms will disappear together.
The same tmth may be demonstrated more elegantly bj the principle
of the greatest common divisor.
Dividing the coefficient of the third term by that of the second, we
getj after two divisions^ a remainder, mQ -f ^ . And, itisevi-
dent that, when this remainder is placed equal to zero, we will have
made manifest the circimistances under which there will be a common
factor between the two coefficients. Placing the remainder eqnal to
Ecro, we get P = -, as before.
XXAMPLES.
1. Make the second and third terms disappear from the equation,
7? — 3a* + 305 — 28 = 0, and find one valne of x,
An$. Transformed equation, y* — 27 = 0, and a; = 4.
2. Find two values in the equation, a^ -|- 4fl^ + ^ + 4x — 15 s Ol
Ant, Transformed equation, ^ — 16 ^ 0, then, y == rk 2, and x «
+ 1, or — 8.
Sixth Proper^,
459. Eveiy equation having the coefficient of the first term plus
unity, and the other coefficients entire, will have entire numbers only
for its rational values.
Let the proposed equation be
a- + Px--' + Qaj»-* + T« + U = 0.
In which, P, Q, &c., are whole numbers.
If X can have a fractional value in this equation, let this vahie.
duced to its lowest terms, be -=-. Substituting this value for a; in the
given equation, it becomes
Multiplying by 1"^^ and transposing, we get
y«— Pa— » — Qa"^6 — Toi^ — U6^«.
OINSBAL THXOBY OF EQUATIONS. 447
Bat the seoond member is an entire quantity, since all its terms are
entire. Hence, the first member mnst be entire. But, since a, bj
hjpothesb, is prime with respect to ft, a" mnst also be prime with re-
spect to 6. The supposition of the proposed equation containing a ra-
tional fractional value, has then resulted in the absurdity of making an
entire quantity equal to an irreducible fraction. We conclude, there-
fore, that this supposition is wrong, and that the rational values are all
entire. The demonstration is restricted to rational values, because the
assumed value, ^, is rational in its form.
o
CoroUary.
460. Articles 452, 456 and 459, are used in solving numerical and
literal equations by changing their forms.
Let the proposed equation be y* — 4y' — 8^ + ^2 =» 0. We know
that all the rational values are entire (Art. 459), and that they must be
found among the divisors of the last teim (Art 452). But as there
are so many divisors of the last term, it will be more convenient to
employ (Art 456) to transform the equation into another, whose last
term admits of fewer divisors. Make a? =y — \~Z"/ — y+^* ^®
transformed equation in re, is o:^ — 6a:' — 16a; + 21 &= 0, and the
divisors of the last term, are + 1, + 8, + 7, + 21, and — 1, — 8,
— 7, — 21. On trial of these divisors, we find that + 1 and + 8,
will satisfy the transformed equation, and, consequently, are values.
Hence, x = y + 1 = 2 and 4, in the given equation. Dividing the
first member of the ^ven equation by (x — 2) (Art. 449), it will be
reduced to a cubic equation, and again^ dividing by (x — 4), it will be
reduced to a quadratic, which can be solved.
Seventh Property.
461. Imaginary values enter equations hy pairs.
We are to show that if a-\-hs/ — 1 is a value in the equation, cc" +
P«— • 4- Qx^ -h Tx + U « 0, a. — hy/^^ will also be a
value.
The truth of the proposition is an evident consequence of Art. 451,
for TJ, the last term, is the product of all the values, and it has been
assumed real. But it can only be real, when the imaginary values (if
448 GENEBAL THEORY OV EQUATIONS.
any), enter by pairs, and of snch a form as to give a real product when
multiplied together. Thus, if we have two imaginary values, m +
^ — n and ^ — 4, we must also have two other values, i» — ^ — ii
and — y/ — 4, otherwise, U will not be real. Or, we may demonBtate
the property otherwise, thus: by substituting the assumed value,
a + hy/ — 1, in the given equation, we will have, (a + h^/ — 1)" + P
(a 4- hy/^^y^"^ + , T (a + fty/^^) + U = 0. When
these terms are expanded, the odd powers of ftv' — 1 will be imaginary,
and the even, real. Representing by A all the real terms involving a
or 6, and by B^ — 1 all the imaginary terms, we will have A +
Bv^ — 1 = 0. Now, since a + h^ — 1 is, by hypothesis, a value, the
last equation must be satisfied, but this can only be so when A = 0, and
B^ — 1 = 0, since imaginary terms cannot be cancelled by real ones.
If a — hy/ — 1 be substituted in the equation, the expanded results
will be precisely the same, except that the odd powers of h^ — 1 will
be negative. Hence, we will have A — B^ — 1^0. But in order
that a + 5v^ — 1 should be a value, we have seen that A = and
B s 0. Hence, the equation, A — B^ — 1 = 0, is satisfied, and that
being so, a — hy/ — 1 must be a value.
The absurdity of the hypothesis of a single imaginary value may be
illustrated by an example.
Let us assume that the three values of an equation of the third
degree are + a, +1, and + ^ — 1. The equation must then be
(g — g) {x —X) (x —y/— 1) = 0, or (x — a) (a:* — (1 — y^I^T)
a+v^— 1)=:0. Hence,cB — a=: 0,anda;' — (l+\/ — l>+v'— 1
1 f ? 1
= 0. The second equation, when solved, will give x = — —-^
And we se e that we do not get back again the two values, + 1, and
Remarkt,
462. 1. The product of imaginary values is always positive, and,
therefore, the absolute term of an equation, whose values ar« all imagi-
nary, must be positive.
2. If the second term of an equation containing only imaginair
values, is wanting, these values will all be of the form, dtiy/ — h.
3. Every equation of an odd degree has at least one real value for x^
GBNXBAIi THIOBT OV BQUATIONS. 449
and the Bign of this tbIbo irill be eontiaxy to that of the laet term of
the eqoBtion.
4. Every equatioii of an even degree, whose last term is negative^
has at least two real Tshies for x; one positiyey and one negative.
BXAMPLKS.
1. One value of a; is 4 + y/ — 10| what mnst a second value be?
An$. 4— s/^^^Tio:
2. Two of the values of an equation of the fourth degree are + a,
and — - h, and the last term is — m. How are the remaining values 1
Afu. Beal.
Eighth Property.
468. If the real values of an equation^ taken in the order of their
magnitudes; be a, h, c, d and e; a being }> &; & ^ c, &c. ; then^ if a
series of quantities^ afy h', </, (f , &c. ; a' taken greater than a, ^ be-
tween a and h, <f between b and c; be substituted for x in the pro-
posed equation, the results will be alternately positive and negative.
For, since a, b, c, dy &g., are assumed to be values, the proposed
equation can be put under the form of (x — a) (x^ h) (x'^c)(x — d)
• • > > &C. z= V.
Substituting, for x, the proposed series of quantities, a', b', </, d^, &c.,
we get the following results.
(a' — a) (a' — b) (a' — c) (of — cf) &c, = + result, since all the
fiictors are positive.
(5' — a) (b' — h)(I/— c) (V — d) &c., = — result, since first fSwtor
only is negative.
(</ — a) (</ — 5) (jif — c) ((/ — dy&c, = + result, since first two
&ctors only are negative.
(cf — a) (cP — b) (c^ — c) (cH — d) &c., = — result, since first
ihree factors only are negative
Remarks.
1. We see that, between the quantities, a' and b', which give the first
plus result and the first minus result, there is one value, a. And be-
tween the quantities, a' and d^, which give the first plus result and the
second minus result, there are three values, a, b and c. And, as the
88* 2d
450 QEMSBAL THXOBT OV BQUATI0N8.
same is eyidentl j trae for any number of odd yafaieBy we oooohide dia4|
if two quantities be saccessively substituted for x in an equation, and
give results, with different signs, between these quantitieSi there will
be one, three, five, or some odd numbers of real value.
2. If any quantity, m, and every quantity greater than m, give re-
sults all positive, then m is greater than the greatest value of x in the
equation.
8. If the results obtained by substituting two quantities have the
same sign, then, between these quantities there are two, four, biz, or
some even number of values, or no value at all. Thus, between c^ and
tff which give + results, there are two values, a and b.
Scholium,
464. The preceding property may be demonstrated differently, by
employing a principle of frequent application in all branches of mathe-
matics, viz., that a quantity changes its sign in passing through zero
and infinity. Let the quantity be x =& a; suppose x^ a, the expres-
sion is positive ; it is zero when a; = a, and negative when x -^ a.
Hence, the quantity x — a, changes its sign in passing through zeza
+ M
Again, take , This quantity is positive when x^a; it is in-
X • CL
finite when x == a ; and negative when x^a. Now, in the results
of Art. 463, between the first + result and the first — result, the first
•member of the proposed equation must have passed through zero, and
consequently through a value. In like manner, between the fizst
— result, and the second ^ result, the first member must again have
passed through zero, and consequently through a second value. Then
two values must have been passed through between the first two + re-
sults, &c. By continuing the same course of reasoning, we might
readily show that three values must be passed through between the fast
positive and second negative result ; five values between the first posi-
tive and third negative result; two values between the first two n^ative
results; four between the first and third negative results, &c., &c.|
Limit of the Values, — Ninth Property.
465. If we substitute for the unknown quantity the greatest coeffi-
cient plus unity, the first term of the equation will be greater than the
sum of all the other terms, provided we always affect the greatest co-
efficient with the positive sign.
^ENSEAL THSOBY OF BQUATIONS. 461
Let us resume the equation^
X* + Px»-» + Qa;"-* + Ta; + U = 0.
It is plain that the most unfavourable case will be when all the co-
efficients have the same sign, + or — . We will assume all the coeffi-
cients after the first to be negative ; the given equation will then be of
theform, «-— Pa:— * — Pa*-* — Px**-* — Pa*-« — pt=0,
or X- — P(a— » — a*-« — a;"-^ — 1) = 0.
Disregarding the sign of P for a moment, we wish to ascertain what
value X must have, in order that of > P(«^^* + ^x^ + af*^ , . , +x
+ 1). Inverting the order of the terms within the parenthesia; we
will have a geometrical progression, whereof, 1- is the first term, x***'
the last term, and x the ratio of the progression. The formula, 8' ai
=-, will then give the value of the quantity within the parenthesis, as
a* _ 1 ^x^ 1\ Px" P
=-. Hence, we have x" "> P ( =- 1, or x" ^ =• =-.
X — 1 ' ^ \x — 1/' ^ X — 1 X — 1
Px"*
This inequality will evidently be true, when x" = =^. Dividing
both m^nbers of this equation by x", and clearing of fractions, we find
a; = P + 1, as enunciated.
It is to be observed that, in this demonstration, we assume x ^ 1.
EXAMPLES.
1. What number substituted for x in the equation, oi^ — 7a^ + ^ +
5x' -f X = 0, will make the first term gre^r than the sum of all the
other terms ? Am, 8.
2. What in the equation, x»— 10x» -f 12x + 18 = ?
Anz, 14.
8. What quantity in the equation, 7? — 2ax" + 6ax — 12a = 0.
Am, 1 + 12a.
In seeking for the positive values of an equation, we need not go
beyond the greatest coefficient with a positive sign prefixed to it, plus
unity. Thus, in^ example first, there is no positive value greater than
8^ because 8, and all numbers greater than 8, will continually make the
first member positive (Art. 468).
462 OXNSBAIi THJlOaT Of IQVATI0N8.
Second Limit, — Tenih Property,
466, If we substittite for x in the equation, a" + Po"^' +Qa^
• . . . Tx + U 3= 0| nni^, increased by that root of the greatest negi-
tiye coefficient which is indicated by the number of terms preeedbg
the first negative term^ we will have a superior limit of the poeatzre
Talues.
Suppose Nx"** to be the first negative term, and suppose W tobe
the greatest negative coefficient. The most unfavourable case obvioaslj
is that in which we suppose all the coefficients after N, negative, and
equal to W. It is plain, moreover, that any value of x which makes
«■ > W(x"*^ + a;""^' 4- fic + 1), will, of course, make the
first member of the given equation positive. We are to find the tbIqb
which fulfils this condition.
Resuming the inequality, x" > W(a:"~" + af"^* + aT ■ * ....
. + « + 1), we have, also, aj"]> W( —^ — 1, or af ^
r. And this inequality will be true, when a;" ^ =-.
X — 1 ^ -^ ' X — 1
W
Prom which, a: — 1 = -^, or (x — l)a;'^* «= W. Now, if (x— 1)
(x — 1)"-' = W, then (x — l)a:^' will be greater than W. If, how-
ever, we place it equal to W, we wiD have (a? — 1) (x — l)"^ = V,
or (x — !)■ =s W. Hence, a? s= 1 + yW, which agrees with the
enunciation. •
It will be seen that in two respects we have taken an un&vorable
case. The second limit t en may considerably exoeed the greatest
positive value, but it is smaller than the first limit, and, therefore, the
most used in practice.
1. Bequired superior positive limit of the values in the equatioO}
«» + 2«« — 8x» + 7a:« — 17x-f5 = 0. Ans. I +s/Vr
2, Bequired superior positive limit of the values in the equation,
«* + 4x* + 12a;«— 5a — 16 = 0. Am, 1 + yiS^
GXNS&AIi THIOBT OF XQVATIONB. 468
Eleventh Propeiiy.
467. The equation, a* + Px*-'+ Qc^*+ + Tx+ II = 0,
oan be transformed into another in y^ or some other variable; snch, that
the yalues of the new unknown quantity shall be some multiple of
those of the old unknown quantity.
For, let y == nx^ then x = ^ . Substituting for x^ wherever it ocoun
in tihe ^ven equation, its value in terms of y, w« have ^ + P^sn +
€11.
Ty + n"XJ a 0. And we see that the transformation is effeeted
by changing x into y, multiplying the coefficient of the second term by
the multiple n, that of the third torm by n', &c. For instance, let it
be required to transform the equation, a^ + 2x' + 4a:' + 8j;-f- 1 = 0,
into another, in which the values of the new unknown quantity shall
be twice as great as those of the old. Then y ^2xj and we havey*
+ 4y» + W+ 6^ + 18 = 0.
Q^+ T?^ + U = 0, or y- + nPjT^ + n^Qy-^
Remarki,
The piineipal use of this transformation is in clearing an equation
of firaotional coefficients, and at the same time making the new equa-
tion preserve the proposed form; thatisjtheeoefficient of the first term
is still to be plus unity. Let us take tSe equation, o^-f h
( V^ • Raf^ Tx TJ
■ -^ 1 — ss 0, in which we assume mn to be
fi fMi n n
the least common multiple of the denominators. Then y a mnx.
V* Pi/*'' Qy«'<
The transformed equation will then be .^ H — ^ x»»i +
Ry^ T^^+n = 0, or y- + nPjT-' + mSiQjr^ +
. •
wSiTly"-* .... m"-'n"-*Ty + m"»""'U = 0. And we see that the
transformation is effected by chan^ng x into y, and multiplying the
second term by the first power of the least common multiple, the third
term by the second power of that multiple, Ac. Thus, the transfoimed
equation of a:' + ^^|+ 1 « 0, isy»+ 4y« — 36y + 1728 « 0.
454 GENERAL THEORT OF EQUATIONS.
468. The two rales just given are, of coarse, only applicable wliea
tlie coefficient of the highest power of x is plus unity. When that is
not the case, we may either first make this coefficient plus unity, and
apply one of the preceding rules, or we may at once make y = fix, n
in this case representing the product of the least common multiple of
the denominators hy the coefficient of the highest power of x.
EXAMPLES.
1. Transform the equation, 9? — 4a!:' + 2x + 2 &= 0, into another,
in which the values shall he twice as great as in the given equation.
Afu. y* — 8t/»+ 8y -h 16 = 0.
2. Transform the equation, a^ — 8a:* -f 2x' + ^ + 1 = 0, into an«
other, whose values are trehle those of x.
Arts, y* — V + 18^ + 27y + 81 =0.
3/ ^Ju X
8, Transform the equation, x* — ■o- + 'q- + c + l = 0, into an
other, containing only entire coefficients.
Am, y^ — 15y> + 300y« + 5400y + 810000 = 0.
4. Transform the equation, 2x' — -^-f ^ + 1 sb 0, into another,
o O
whose coefficients shall all be entire.
An», 2y»— 2y» + 6y + 216 ^ 0,
tXj X
5. Transform the equation, 2x* — "o'+fi'^l^O* ^*o another,
whose coefficients shall all be entire, and the coefficient of the fint
term, plus unity. Ans. y* — 2y« + 12y + 864 « 0.
In this, make y = 12x.
469. When the coefficient of the first term is some whole number
different from unity, and the other coefficients are entire, make n, in
the equation y = nx, equal to the coefficient of the first tenn. Thus, to
transform the equation, 6x' — 19x*+ 19x — 6 » 0, make y = 6x,
the transformed equation will be y" — 19y" + 114y — 216 =« 0.
Tweifik Property, — Process of Divisors.
470. In every equation in which the coefficient of the first tenn is
unity, and all the other coefficients are entire, a value will divide the
List term, the sum arising from adding the quotient so obtained to the
OINSBAL THBOBT OW XQUATIONS. 455
ooei&cient of tilie seoond tenn from the liglit, the sum arisiDg from
adding this second quotient to the coefficient of the third term from the
right, and will thus continue to be an exact divisor of all the successive
sums so formed, until the coefficient of the second term from the left
has been added to the previous quotient^ when the last quotient will be
minus unity.
Let us resume the equation,
»- + Px--' + Qa;*-* ....... + Tx + U = 0,
in which all the coefficients are entire. Suppose a to be a value, the
equation will then become, after transposition and division bj a, a""*
+ Pa*^ + Qa^ T = — ' — . Now, since the first member is
a
made up of entire terms, the second member must be entire also.
Hence, U is divisible by a, as it ought to be, since U is the product of
all the values. Transposiii^ T to the second member, we have a"~* +
Pa*^ + Qa"^ + &c. sss — T a= U'; U' representing the alge-
braic sum of T and — .
a
Dividing both members again by a, we get the equation, a"*"*-!-
P(^^ + Qa-^ + 4o. = — ^.
d
The first member being entire, the second member must be entire
also. Hence, the second condition to be fulfilled by a value is, that it
must be an exact divisor of the sum formed by adding the quotient of
the last term by the value to the coefficient of the first power of x.
This was also to have been anticipated, because the coefficient of the
first power of a; is made up of the values, taken m — - 1 and m — 1 ;
one of these combinations will not contain the value a, and will have a
sign contrary to the first quotient, and will be cancelled by it. The re-
maining terms making up this coefficient all contain a, hence, a will be
a divisor. By transposing the terms in succession, and continuing the
F
division, we will find, after m transpositions and divisions, — = — 1 ;
F' representing the sum of the coefficients of x*~' added to the pre-
ceding quotient.
Hence, the last condition to be fulfilled by a value is, that it must be
an exact divisor of the sum formed by adding the coefficient of x"^' to
tlie preceding quotient.
456 OINIBAL THXOBT OF EQUATIONS.
We, iherefore, have the following role for finding the rational values
of an equation.
RULE.
I. Transform the equation, if necessar^y trUo another, in tehi^ att
the coeffkienti shall he entire; that of the first term being j^u$ tintle^.
All the rational values vnU then he entire (Art. 459).
II. Find the siufperioT positive and superior negative limits of the
values,
m. Write down in succession, in the same horizontal line, all the
entire divisors of the last term between zero and these limits,
lY. Divide the last term by each qf these divisors, and write the r^
tpective guotients beneath the corresponding divisors,
Y. Add the coejfficient of the first power of z to each of these quo-
tients, and write each sum thus formed beneath the corresponding
quotient.
YI. Divide these sums by the corresponding divisors in the cohmsm
of divisors, and write the new quotients under the corresponding wusnSf
refecting those divisors which give fra/Aional quotients, and crossing osU
the vertical odlumn in which they occur,
VJJL. Proceed in this way untU the coefficient of z"^"* has been added
to the preceding quotient, then those divisors that give mmtit wsii^for
quotients, are values, and they are the only rational values,
m
EXAMPLES.
a* — 2a* — 6x + 6^0. _
The superior positiye limit is 1 +}/b = 6.
The superior negative limit is found by changing + x mto^^x; the
8aperio]r positive limit of the transformed equation will then be the
superior negative limit of the given equation*
Changing + x into — a, we get — x* — 20^ -f Sx + 6 ss 0. But
the coefficient of the first term must always he plus unity, henee, hj
multiplying the equation by minus unity we havCi a? -f 22? — Six—
6 = 0. _
Then — (1 + \/6) = superior negative limit. Assume — 4 to be
this limit, the square root of 6 being between 2 and 3, we take 3 to be
the root, nnoe it is better to have the fimit too great than too bbsIL
Hence, we have these divisors below the limits, +1, + 2, + 3; — l,
— 2,-8.
OBNXBAL THXO&T OF EQUATIONS
467
PMliiye.
DiyiflOM, + 1, + 2, + 3,
Qaotients^ + 6; + 8, + 2,
Smns, +1,-2,-3,
Qaolients, +1, — 1, — 1,
Sums, ^ 1, — 3, — 3,
Quotients, — 1, x ; — 1.
- 1,-2,
- 6, -8,
-11,-8,
+ 11, + 4,
+ », +2,
- 9,-1.
Hence, we Lave + 1 , + 3, and — 2, for the three yalnes of the
given equation.
Let ns take, as a second example, the equation, x^ — 2x' — 8 » 0.
Superior positive limit, sss + 9; superior negative limit, ==:•— 9.
The equation may be written x* zfc Ox* — 2a:^ rb Ox — 8 = 0.
PDiitlTe.
Divisors, +1, 2, + 4, + 8,
Quotients, — 8, — 4, — 2, — 1,
Sums, — 8, —4, —2,-1,
Quotients, — 8, — 2, X , X ,
Sums, —10, — 4, X, X,
Quotients, — 10, — 2, X , X ,
Sums, -10,-2, X , X ,
Quotients, — 10, — 1, X , X ,
NegatiTt.
- 1,-2,-4,-8,
+ 8, + 4, + 2, + 1,
+ 8, + 4, + 2, + 1,
- 8,-2,
-10,-4,
+ 10, + 2,
+ 10, + 2,
- 10. - 1,
X,
X,
X,
X, X,
Heaee, + 2 and — 2 are Talaes in the ^ven equation. And by
dividing the fiist member of the ^ren equation by the factors (x — 2)
and (x + 2) corresponding to theee valuee, we obtain the quotient,
a^-f 2. By placing this quotient equal to zero, the remaining two
values of the equation will be found to be + y/ — 2, and — y/ — 2.
471. The process of diTisors is applicable to literal, as well as nu-
merical equations, when all the coefficients are entire. A literal equa-
tion, too, will beet show why it b that the successiye sums are divisible
by the values.
Take, as an example,
a? — (a +6 + c)7? -f (o6 ■{■ ac+ bc)x — aic i^ 0.
Since the terms are alternately plus and minus, the values must all
be positive.
89
468
GBNSBAL THEOET Of EQUATIONS.
Biyisora^ + a, + 6, + c.
QaotientSy — 6c, — ac, — ah.
SrnnB, ah + ac, ah + he, ac + he*
Quotients, h + e, a + e, a + h.
Sums, — a, — hj — c.
Quotients; — 1, — 1, — 1.
Hence, the three Talues are + a, + &, + e.
We will take, as a second example in literal eqaationa,
ct*^. (a — 6 + 2)x«+ ( 2(a — 6) — a5)a5 — 2a6 = 0.
FOSitlTtt.
Diyisors, + a, + 6, + 2,
Quotients, — 26, — 2a, — ah,
Sums, (2a— 6a — 46), — 6(a + 2),
2(a — 6 — a6),
Quotients, x, — (a+2,)(a — h—ah\
Sums, X,— 6, 2(a+l)— 6(a + l),
Quotient, X , — 1, x.
NegftttTe.
— a, — 6, — 2,
+ 26, + 2a, + a6,
a(2 — 6), 4a — 6(2 + o), 2(a— 6),
+ (6-2),x,-(a-6),
+ a, X, +2,
— 1, X,— 1.
Hence, the three values are + 6, — a and — 2.
The process of divisors being of such high piactical importance, we
will give another example, in which one of the sums is £ero«
Take the equation,
re* — «■— 4x + 4 = 0. •
Superior positive limit, as 5 ; superior negative limit, es -— 8.
PO«itlT«.
Divisors, + 1, + 2, + 4,
Quotients, + 4, + 2, + 1,
Sums, 0, — 2, — 8,
Quotients, 0, — 1, X ,
Sums, — 1, — 2, X,
Quotients, —1,-1, X ,
NcgatlTe.
-1,-2,
-4,-2,
-8,-6,
+ 8, + 8,
+ 7, + 2,
-7,-1.
Hence, the three values are + 1, + 2, and — 2.
472. The process of divisors enables us to find all the rational vahies
of any equation. We have only to make all the coefficients entire, if
not already so, that of the highest power of x being made plus unity.
All the rational values of y in the transformed equation will be entire ;
find these values, and then those of x will be known from the relation
between x and y.
Take 6x«— 19x« + 19x — 6 = 0. Making y = 6x, we have tbe
QBNBBAL THBOBT OV BQUATI0N8. 459
transfoimed equation in y, y' — 19^ + 114y — 216 s 0. By pro-
ceeding as before, we find the three yalnes of y to be + 9, +6, and
+ 4. HenoO; the corresponding yalnes of a; are | + 1 and + |.
478. In thisy and in many examples; the last term has a great num-
ber of exact diyisorS; and the process is therefore tedions. It is often
convenient to transform an equation, whose last term is too large, into
another whose values shall be greater or smaller by a constant quantity.
Making y c= z + 2, in the preceding equation, the transformed equa-
tion in 2 is s^ — 132^ + 50z — 56 sb ; and the three values of z
are 7, 4, and 2. Hence, those of y are 9, 6, and 4, and those of x,
-f I, + 1, and + I, as before.
474. If we solve the equation, a^ — 2x* — &a^ + 8x — 4 = 0, by
the process of divisors, we will get the three values, +2, — 2, and
— 1. And, by dividing the first member of the equation by the factors
corresponding to these values, we will have left, a; + 1 » 0. Hence,
the value — 1 enters twice in the given equation. We see, from this
example, tiiat while the process of divisors gives the rational values, it
does not show whether any or all of them are repeated once or more.
This remark is evidently applicable to all equations whatever. Some
test is then necessary, by which we can ascertain the equal values.
EQUAL VALUES.
475. Let the equation be
(x — a)-(a — 6)'(x — cy (x— <0(a? — c) &c. « (A),
in which there are m values equal to a, n values equal to 6, j9 values
equal to c, and all the other values unequal. Calling D the differential
coefficient of this equation, we will have (Art. 866),
B ^m(x — a)—' (x — by (x — e^ (x — d) (x — e) &c. +
n(x — fc)*"* (x — a)" (x — cy (x — d) (x — c) &c. +jp(x — o)'-'
(05 — a)" (x — h\ (x — d) (X — c)&c. + (x— a)- (x — 5)» (x — cy
(x — «) &c. + (x — a)" (x — by (x — cy (x — d) &c., plus other
terms (B).
It is plain that the greatest common divisor, IX, between (A) and
(B), will be
F = (x — a)—* (x — ft)-' (x — cy-\
We see that this divisor contains (m — 1) values equal to a, (n — 1)
values equal to b, and (p — 1) values equal to c. Hence, it is plain
that the number of equal values in each set of the greatest common
460 OEKBBAL THSORT OV SQ17ATIOK6.
divisor is one less than in the giy equation. Therefore, to aseeitnn
whether there are equal valnes, we have this
RULE.
Find the differential coefficient of thefirtt member of the (/iven equo'
turn {or J as it is generally called j the first derived polynomial), next get
the greatest common divisor hetween ^ first member of the propomd
equation and this differential coefficient, place this common divisor equal
to zero, and find its values. Each independent value so foumd vnXJL he
repeated once ofitener in the given equation than in the greatest commtm
divisor; if the loiter , for instance, contain two values equal to sl, and
four values equal to h, the former vnll contain three values equal to a,
and five equal to b. If the greatest common divisor he of too hi^ a
degree to solve, toe may find the second differential coefficient, or second
derived polynomial of the given equation, then the greatest eomflmas
divisor hetween this polynomial and the first member of the given ejnei-
tion. Each independent value of this common divisor will he r^
peated twice ofUner in the given equation.
EXAlfPLES.
1. Does the equation^ nc^ — 2a^ + fx' — x+ } = Oj oontain eqad
values f
First derived polynomial, 4a5* — 6x" + 3a5 — 1.
Greatest common divisor^ x — 1, which, placed equal to sero^ gives
x = l. Hence, the given equation contains two values equal to L
If we divide that equation by (x — 1)*, we will have left, a? + } = 0.
Hence, the other two values are +v/ — }, and — \/ — J.
2. Does the equation, a:* — a?* — 14x* — 2fer' — 19x — 5 = 0,coii-
tain equal values f
First derived polynomial, 5x* — 433* — 42a^ — b2x — 19.
Greatest common divisor, (x + 1)'. Hence, the given equation eon-
tains four values equal to — 1, and, by dividing the equation by (x + Vf,
we find the other value to be -f- 5.
8. Required the equal values in the equation, a^ — 2af +1 = 0.
Ans. Two values =s + 1 and two values = — L
QBMXBAL THBORT OF IQUATIOITB. 461
4. Beqnured all the Talues of the equation^ t? — ~ — 2as* + ^ + ^
— J = 0. Am. + 1, + 1, + }, — land — 1.
5. Beqiiired all the values of the equation, ^ -I- a^ -*- M *— - 4a;
4- 8 «. 0. Am. + 1, + 2, —2, —2.
After diyiding out the equal factors, (x + 2) (x + 2), we had a
quadiatio left, which was solved by known rules.
6. Find the values of the equation, 7^ — aj* — 2a? + 2x" + a5 — 1
= 0. Ans. +1, +1, +1, — 1, and — 1.
The first greatest common divisor is a:* — 7? — se + l. As this,
when placed equal to zero, is too high to solve, we find the 'greatest
common divisor between the second differential coefficient of the given
equation, and the first member of that equation, or, what amounts to
the same thing, tho greatest common divisor between x" — x*— x4- 1,
and its derived polynomial, Sx* — 2x — 1. This common divisor is
(x — 1)'. Hence, the given equation contains three values equal to
plus one, and, by dividing out by the factors, (x — 1) (x — 1) (x — 1),
or, X* — 3x" 4- 8x — 1, we have left, x* + 2x + 1 = 0. Hence, the
other two values are, — 1 and — 1.
7. Solve the equation, x* + x* — ^ — «" + 8x — 4 « 0.
Am. X = + 1, +1, + 1, — 2, — 2.
8. Find the five values of the equation, x* + 2x* — 16 x — 32 = >
Ans. + 2, — 2, — 2, +%/— 4, — v'— 4.
9. Find the values of the equation, x* + x* — 2x^ — 16x^ — 16x +
32 = 0. Am. + 1, + 2, — 2, — 2, +^'"=17—^— *'
10. Find the values of the equation, x»— (2a + 5)x*+ (a«+ 2a6>
_ a^h = 0. Ans. + a, + o, + ^•
11. Solve the equation, x* — 2 (a + 6)x» + (a« + 4a6 + h'^af — 2
(a^b + aV)x + a^b^ = 0. Am. + a, + a, + ^> + ^
39*
46B GINIEAL VHXOET OF XQUATZOWB.
DERIVED POLTNQMIALS.
476. We have had ocoafiion to use the relatioD, ^ = a 4- y^ in tni»-
fonniiig the general equation in x into another in y, saoh, that the
second term was wanting. But the transformation in terms of y and a
is of more general application. Sometimes a is an undetermined oon-
stant, whose value is afterwards determined in such a manner as to
make the equation fulfil a given condition. Sometimes a is a deter-
minate numher or quantity, which expresses the di£ference between
the values of the given equation and another which we wish to fonn.
Suppose, for instance, we wished to transform the equation, a^ — x^ -^
4a;+ 5ss0, into another in y, such, that each value of x should exceed
hy 2 e^h corresponding value of y. Then, a must be made equal to
2 in the relation, x ^ a -^ y, and we must substitute 2 -f- ^ for x
wherever it occurs. Hence, the equation in y would be (2 + y)* —
(2 4- y)' + 4(2 + y) + 5 s= 0, or, expanding and reducing, y* + 5y*
+ 12y + 17 B 0. In this case, the transformation is eadly effected
by actually substituting the value of x in terms of y, and developing
the several binomials in the new equation. But, when the given equa-
tion is of a high degree, this process is tedious and impracticable.
Suppose, for example, it were required to transform the equation, x" —
40a;" + x««— 2x» + 8x«+ 5x^ — 4x* + Ox* + 12«^ — «• + 3x + 15
tsB 0, into another in y, such, that the values of x should be less by
unity than the corresponding values of y. Then, a as — 1 in the re-
lation, x = a + y, and we must substitute for x, wherever it ooenn,
(y — 1). It is plain that the development of the several binomials in
the new equation will be exceedingly tedious.
The method of derived polynomtah enables vs to effect the required
&ans/ormatton in y toithaut svhstitution and devdopmewL
Let us resume the equation,
X- + Px-^-' -f Qx--« -f- Tx -HJ « 0,
and let us suppose, as before, x=ia'\'y.
The new equation in y will then be
(a + yr+ P(a + y)"-' + Q(a + y)"^ T(a + y).h XJ = 0.
Let us assume
(a + y)- + P(a +y)—' + QCa + y)"^ T(a + y) + U - A
+ By + Cy* + Dy*+ Ey* + &c.
QBNXBAL THBORT OT EQUATIONS. 468
Making yssO, we get
a- + Pa—* + Qa-* + Ta + U = A.
Differentiating the given equation^ and dividing by dy, we get
K« +y)""' + P(»» — 1) (« + y)"^ + Q(«* — 2) (a + y)^
+ T = B + 2Cy + 3iy + 4Ey (R).
Now, make y = 0, and denote by A' what B becomes in that oase^
we get
«Mi— » + (m— 1) Pa— « + (m— 2) QaT* + T = B = A'.
Differentiating (R), we get
m(«» — 1) (a + y)— •+ (i» — 1) (w — 2) P(a +y)— •+ (m — 2)
(w— 3) Q(a+ y)-» + &o. =20 + 2 . 3Dly + 3 . 4iy + Ac. (8).
Making y = 0, and representing by A" what 2C becomes, we get
m(m — l) ar^ + (m — 1) (w— 2) Pa— *+ (m — 2) (m— 8) Qa— •
+ &Q. = A" « 20.
Then C^^" ^ ^(^'^~^) a— *+(w — 1) (m — 2)Po*-^+ Ac.
2 2
Differentiating (S), we get
fn(m — 1) (m — 2) (a + y)"^+ (*» — 1) (m — 2) (w— 3) P(a+y)— •
+ (w — 2) (m — 8) (m — 4) Q(a + y)— * + &o. » 2 . 8 . D +
2.3. 4Ey + &c.
Making y »= 0, and preserving a similar notation, we have
_ A^^ m(wir-l) (m— 2) g— ^+(m— 1) (m— 2) (m— 3) Pa—
^""2.3*^ ^ 2TS
(m-2)(m-3)(m-4)Qa-*
By proceeding in the same way, all the other coefficients ooold be
determined. But the law of formation is already apparent, A is what
the given eqnadon becomes when a (or the difference between x andy)
is substituted for x; A' is formed from A, by multiplying the coeffi*
dent of a in every term by its exponent, and then diminishing that
A" .
exponent by unity; -^ is fonned in precisely the same way, except
A'"
that the result is divided by 2 ; -»- is fonned in like manner from A",
except that the result is divided by 3.
A""
-^ is formed according to the same law from A"', the result being
divided by 4.
464
GKNSBAL THBOBT Of KQUATIORS.
II
P
+
+
+
§
J
§
1
+
+
+ I
+ +
+
+
+
\
+
I
+
00
I
ca
I ^
C4
1.
I
g
s
+
+
+
I
I
CO
I
I
+
+
I
I
1
CO
i
s
C4
^ 1
I
t
04
CO
04
I
Pm
1
•I
T
s
^»-/
1
f
1
1
1
CO
JS
u
a
o
s
+
+
>
CO
09
+
00
>
09
^
09
^^
CO
Gi
5
+
09
+
I
I
Q>
+
i
^
^
a
I
9
i"
.s
.2
I
•s
§
4
.•^ a s i
J .fc •§ -S
9 'O w« "**
P^ 4a 2
s <S I ;§
® ® " -»
<=> ^ ^ ^
II
GINX&AL THBOBT OV EQUATIONS. 465
477. To transform an equation into another, in wUch the Yalnes of
the new unknown quantity shall differ from those of the old by some
constant quantity^ we have then this
BULE.
I. Replcbce z in the given eguatiofiy wherever it occurZj by the assumed
difference^ and cajl the result A.
n. Find the derived polynomial of A, and caU it A'.
m. Find the derived polynomial of hl^ and call it A".
rV. Find all the other derived polynomials in the same^ and calcu-
late their values,
Y. Substitute the vahtes so found in the formula, A + A'y +
A'V A"y
-g ti + = — 5 — s- + &c. as 0, and the transformed equation toUl be
found.
1. Transform the equation, a? — a^ + 4a; + 5&=0, into another in
y, 80 that the yalnes of x shall exceed those of y by 2.
Then we have, by the rule,
(2)»_(2)» + 4(2) + 5 = 17 = A,
3 (2)« — 2 (2) +4 c= 12 « A/
6(2) — 2 =ilO«A",
6 = 6 = A'",
» 0=:A»^
A'V A'^v*
Hence, the formula, A + A'y + =-~r + := — —^ + &o. = 0, giyes
17 + 12y + ^^ + y* = ^ or, changing the order of the terms, ^ +
5^ -f 12y + 17 s 0, the same as before found.
2. Transform the equation, x* — Sx* + 4a^ + a? + 1 = 0, into an-
other in y, so that the values of x shall exceed those of y by unity.
(1)- - 5 (ly + 4 (1)» + (1)« + 1 - 2 « A,
6 (1)» — 26 (1)* + 8 (1) + 1 = — 10 = A',
80 (1)* — 100 (ly + 8 =. — 62 = A",
120 (1)» — 300 (ly = — 180 = A'",
360(iy — 600(1) = — 240 = A%
720 (1) » 720 = A^
720 = 720 a. A^».
2e
466 OENBBAL THEOBT OV SQUATIOKS.
Henoe; the transformed equation is
2 — lOy — 3V — 30^— 10y* + 6y»+y« = 0,
or, f/" + 6/— lOy* — 30y*— 3V — lOy +2 = 0.
3. Transform the equation, x^ — 8a^ + Ox* + a; + 12 =: 0^ into an-
other^ whose second term shall be wanting.
Am. y*_15y" — 27y + 2 = 0.
In this example, a; = y — (~|) = y + 2 (Art. 455).
4. Transform the equation, 2a? +a^ + ^ — 4 = 0, into another,
whose second term shall be wanting. Ans. ^ + y^ — ||^g = 0.
First divide the given equation bj 2, to bring it under the proposed
' form. Then, make a = y — ^2 = y — J.
5. Transform the equation, — ix? + 7?-\- x — 1 ssO, into anotfaa*,
whose second term shall be wanting. Am, ^ — |y + ^4 ^^ ^'
First bring the equation under the proposed form by multiplying by
minus unity. Then, make a; = y + i.
6. Transform the equation, a? — 18x" -f 185aJ* + ib* — a^ + 2x+i
=0, into another, whose second term shall be wanting.
Ans. %f + 541j/» + 8653y» + 8771y + 7318 = 0.
Why- did the second and third terms disappear together?
7. Transform the equation, ^ — 7x' -h x* — 2x* + a^ — 3x^ + %
4-8 = 0, into another in y, such, that the values of y shall be less by
unity than those of x.
Am. y^ — 20y»— 67y* — 102y«— 86y»— 40y « 0.
Why is the absolute term wanting in the transformed equation T
Why did the second term disappear ?
8. Transform the equation, 4x^ — Zo^ 4- 2a; -^ 3= 0, into anoilier
in y, such, that the values of y shall be greater by 2 than those of z.
Am, 4y» — 27y"+ 62y — 51 = 0.
It will be seen that + 1 is a value of x in the ^ven equation, and
-f 3 is a value of y in the transformed equation, as it ought to be.
9. Transform the equation, ^ — 3x' + 2x' — 3x =: 0, into anotfier
m y, such, that the values of y shall be less by unity than those of x.
Am, 4y*+ ISy* + 17y"+ lOy = 0.
QXNEBAL THXOBT 07 SQVATIONS. 467
Why IB the absolute term wanting in the transfonned eqiiadon?
Why are all the terms positive f
10. Transform the equation, 5x* — 50x* + 200x' + 9x* + x — 165
^ 0, into another in y^ such, that the yalues of y shall be less by 2
than those of x. An». b/ + 409y" + 1237y + 838 « 0.
Why did the second and third terms disappear ?
EQUATION OF DIFFERENCES.
478. We have now shown how all the rational yalnes conld be ob-
tained, either by operating directly upon the given equation, when of
the proposed form, or upon the transformed equation, and then, by means'
of the relations between the values in the transformed and giveki equa-
tions, ascertain the rational values in the given equation. If we would
next divide the given equation by the factors corresponding to its ra-
tional values, and place the quotient resulting equal to zero, we would
have a new equation containing only irrational and imaginaiy values.
Bei^re explaining the method of finding the irrational values, it becomes
necessary to show how to find an equation, whose values shall be equal
to the difference of values in the given equation. Such an equation is
called the eguation of differences.
Let afy a/', x"', &c., be values in the given equation, and y a value
in the equation of differences sought. It is plain that this equation
will be of the same form, whatever pair of values we assume in the
given equation. Hence, we may assume y = x" — x', then a!" =
y 4- x'. Since x'' is, by hypothesis, a value in the ^ven equation, x*
+ Px—* + Qx"*-" + Tx + U = 0, by substitution, we will
have x''- + Px"-' + Qx"-^ Tx" + U « 0. Noifr, if we
replace cd' by y+ ^ in this equation, and develop by the formula in
Artide 477, the new equation will become Xo 4- X|y -f- Xg ;.^ + Xs
JL . ^
- — ^ — ^ -|- &c. = ; in which X© denotes what the ^ven equation
becomes when x' is substituted for x, X, is the derived polynomial of
Xo; Xg the derived polynomial of X„ &c., &c. But, since ci is, by
hypothesis, a value, Xo is equal to zero, and, by dividing by y, the new
equation becomes X, -|- ^8=-^ + Xi -i 2 ^ + &c. = (D), and is
468 GENERAL THSORT OF EQUATIONS.
the equntion, wbicH oombined with the giyen equatioD, will give the
one sought.
479. It is obvious that, to obtain (D"), we have only to get the deri-
yatiye of the first member of the given equation, add to this the second
derived polynomial, multiplied by the first power of y, divided by 1 . 2,
plus the third derived polynomial, multiplied by y", divided by 1 . 2 . 3,
plus other derived polynomials, multiplied by corresponding powers of
y with their appropriate divisors.
As a simple illustration, let it be required to find the equation of
differences to the equation, a^ — 4 s 0. Then X, &= 2x, Xg ^ 2, and
(D) becomes 2^; -f ^ = 0. Eliminating x between these equations,
by the method of the greatest common divisor, we have,
4a:" — 16 \2x+y
iaf + 2xy 205 — y == quotient.
— 2a;y — 16
— 2yy — y *
y — 16 B* 0, equation of differences.
The values of y in this equation, are + 4 and — 4, and those of x
in the given equation, + 2 and — 2, the difference between which is
either + 4 or — 4.
If in the equation, y" — 16 = 0, we make ^ = a?, we will have a—
16 = 0, and the value of z will be equal to the square of the difference
of the values of x. Such an equation is called the eqtuition of ike
9quare of the differencei.
The degree of the equation of differences will be expressed by the
number of combinations of m values, taken two and two, or by
m (m — 1), and, since this product is always even whatever may be the
Tslue of m, the sought equation will always be of an even degree.
As a second illustration, let it be required to find the equation o€
differences to the equation, a? — x — 6 = 0. Then, X, s= 3x* — 1,
X2 ■= 6a;, Xj = 6, X4, Xj, &c. = 0, and equation (D) becomes 3a::* — 1
+ ^Tcy-^jt = ^} ^h 3x'4- 3ay + y* — 1 = 0; and, preparing the
given equation for division by multiplying by y, we have
GENERAL THEORY OF EQUATIONS. 469
8a* — 8« — U I 8a«+a»y+y^l
aac*-f8a^+gy*— a? »— y=Quotient.
2(y«— l)a; + y»— y— 18
2(y«^l)
6(y*— l)a^+6(y»-l)ry+2(y*-l)« | 2d Qnotle&t =: 3*.
6(y*— l)ai«-h8y*a»— 8ay-4^
8(yt^— l)xy-|-54«+2(y«— 1) =: 9d Seiiudndflr.
or, 8(y*-y+18)«+a(y«-l)* MultiplTliig 1^ 2(y^-l)
2(y»-l)
6Cy» - y + 18) x(y* — 1) + 4(y» — !)• | 2(y*--l>B-|-y*-y— 18
6 (3>'-y-fl8Ky*-lH-3(y'-y-hl8Ky«-y-18) | 8(y«-y+18)=:8d Qwrttalt
4(y»— 1)« — 8 (y«— y + 18) (y« - y — 18) =
or, developing 4(y« — 3y* + 3/ — 1) — 3/ + 3^ + 54y» + 8y* —
8/— 54j^ — My» + 54y + 3(18)« = ; or, reducing 5^— 6y + V
-f- 968 >s 0, wliioh is the equation of differences sought.
Now, make 3/^ = 2, and we have for the equation of the square of
the differences s* — 62" + 9^ + 968 = 0.
In the above development, ihe alternate terms were struck out This
was to have been anticipated from what had been said. But we may
show, in another manner, that the equation of differences can contun
only even powers of y.
For, let a, h, c, d, &c., be the numerical differences of the values of
the given equation, then, + a, — a, + 6, — 5, &o., will be values in the
equations of differences; and, since the factors corresponding to these
values, permuted in pairs, give us (y + a) (y — a), (y + 6) (y — 6), Ac,
the first member of the equation of differences will be, (y" — aF) (t^—tF)
(j^ — ^y &0.
Now, making y* = 2?, we will have (z — a") (2 — ft") (z — c") &o. =
for the equation of the square of the differences. The degree of the
last equation will only be half as great as that of the equation of dif'
ferenoes.
EXAMPLES.
1. Given, a* — 9 = 0, to find the equation of differences.
A'M. y« — SrfrsO.
2. Given, x' + 92; + 4 = 0, to find the equation of differences.
AuB. y» + 54y* + 81^* + 3348 a 0.
After the first division the remainder will be, 2(y' H • 9) re + y" + 9y
+ 12. Multiply the last divisor twice by 2(y» + 9). Then,
after two divisions, you will have 4(y' + 9/ — 3(y* +9y — 12)
^2^ -j- 9y -I- 12), which will reduce to the expression above.
40
470 GENERAL THEOBT OF EQT7ATI0NB.
8. Given; oi?+ ax-{' h ==0, to find tlie equation of differences.
Ans. y" + 6ay* H- 9a« y* + 4a» + 276" = 0.
First remainder is, 2(y' + o') x + y* + ay + 36. Use 2(2/" + a^ twioe
as a multiplier.
4. Given, ic* — a; = 0, to find the equation of differences.
Ans, s^ — 6t/* + 9i/»— 4 = 0.
First remainder is, 2(y* — l)x+y' — y. The coefficient of x is
used twioe as a multiplier.
It will be seen* that + 1 and -^ 1 are values in the equation of dif-
ferences. Dividing by y" — 1, and solving the resulting equation, y* —
6y^ + 4 s= 0, by the rules for binomial equations, we get the four values,
+ 2, — 2, + 1 and — 1. The six values, then, in the equation of dif*
ferences, are, -|- 2,-2, and + 1 and — 1, repeated twice. These ought
to be the values, since those in the given equation are, + 1 and — 1,
the differences between which are those given above.
IRRATIONAL VALUES.
480. An equation, freed from the factors corresponding to the n-
tional values, will contain only irrational or imaginary values, or both
irrational and imaginary values. We can best explain the procesB of
finding the irrational values by an example.
Let us take the equation, a^ — 2x^ — 2 s= 0, to find one podtire
rational value. The superior positive limit is, 1 + 1/2 =bS. Substi-
tuting the natural numbers from up to 3, we find that and 1 give
negative results when substituted for x in the equation, and that 2
gives a positive result Hence, a value of the equation lies between 1
and 2, and I is the entire part of the irrational value.
Now, make x = y + 1 in the given equation, then the new equation
in y will have values less by unity than those in x. Hence, y will con-
tain the decimal part of the value of x. This transformation can be
most reidily effected by the formula of Article 477. We have
(1)* — 2(1)»— 2 « —3 i» A.
5(1)«_6(1)« = _1 = A'.
20(1)» — 12(iy = 8 = A".
60(1)» — 12 = 48 = A'".
120(1)» = 120 = A>^
120 = 120 « A^
GENERAL THEORY 07 EQUATIONS. 471
^^-^ + ^y + r7¥-^m + T7rf74 + i.2.8 4.5
+ &0. = becomes, — 3 — y + 4t/» + 8j/« + 5y* + t/» = ; or,
changing tlie order of terms, y* + 5y* +82/* + 43/" — y — 3 = 0.
If, now, we make z = lOy, then y = ^. The first figure, then, of
the yalue of z, in the transformed equation, will be tenths in the value
of y; and, consequently, tenths in the value of x. The transformed
equation (Article 467) is,
a» + 502^ + 800a» + 40002* _ lOOOOy — 800000 = 0.
S. P. limit =1+ t/300000 « 1 + 23 « 24.
The limit is here too great for any practical use. Substituting, be-
tween and 10, we find a change of sign in the results corresponding
to the substitution of 5 and 6. For 5, we have,
(5)»+50(5)*+800(5)"+4000(5)«— 10000(5)— 800000=— 115625.
And 6 gives,
(6)»+50 (6)*+800 (6y+4000(6)»— 10000(6)— 300000 = + 29376,
Hence, 5 is the entire part of the value of Zy and this corresponds to
five-tenths in y and x. Therefore, 1 . 5 is the approximate value of a;,
to within tenths. To get a nearer approximation, let us transform the
equation in z into another in tr, so that the values of w shall be less by
5 than those of z. Then w will contain the remaining part of the deci-
mal value of z. The equation in V3 is,
vf + 75w* -I- 2050tt^ + 24750tc« + 118125w — 115625 = 0.
i
Making t &» lOtr, or ir =» ^, the transformed equation in t (Art. 467),
will be,
^+750<*+205000^+24750000««+1181250000<— 11562500000 « 0.
On trial, we find that 8 and 9 give results with contrary signs ; hence,
8 is the entire part of the value of t, and corresponds to '8 in 10, and
to .08 in y and x. We have, then, for the approximate value of x,
1-58. The process would be, in all respects, the same, were a negative
Talue to be found, except that we would find the negative limits.
Let us take, as a second example, a^ — lOos* + 6a; + 1 = 0.
We find that and 1 give results .with contraiy signs, and so do 3
472 OSNEBAL THSORT OT EQUATIONS.
and 4. Let ns find ibe decimal part of the value, whose entire put if
3. Make x &= y + 3. The equation in y will be,
yfi 4. 15y4 + 80y» + 18(V + 141y — 8 « 0,
and that in z,
2» + 1502* + SOOOz* + 1800002« + WlOOOOa?— 800000 = 0.
Substituting and 1, we find a change of sign. Hence, is the
tenths of the given equation. Making z =s + to, "we have,
w» + 150w* + 8000ir» + 180000w' + 1410000w — 800000 = 0.
The equation in t, is,
^+1500f*+800000/"+180000000^-|- 14100000000^-80000000000=0.
We find that 5 and 6, when substituted, give a change of sign.
Now, make ^ n t + 5, the equation in $ will be,
«»+175«*+830250«»+192226280««+15960753125«-4899069875— 0.
By changing this into an equation in r, making r =* 10«, we will find a
change between 3 and 4. Hence, 3*053 is the approximate value of £
to within thottsandtJis.
481. In this process we have proceeded upon the hypothecos, that
but one real root lay between two successive integers. To asceitais
whether this is the case, we have only to transform the given equation
in X into another in y, so that the values of y shall be the squaies of
the differences of those of x. Next, find the inferior limit of the posi-
tive values of y. Suppose D' to be this limit ; then, since D* is less
than the least value in the equation of the square of the dififerenoes,
v^D^y or D, will be less than the least difference between the values in
the given equation. Now, if D be "^ 1, it is plain that no real root
will be comprised between two successive integers, and the prooesB
described above can be pursued. A similar course of reasoning can
be applied to I/', the inferior limit of the negative values in the equa-
tion of the squares of the differences.
But, if D <[ 1, then two or more real roots may be comprised be-
tween two consecutive integers. In this case, we have only to substi-
tute a series of numbers, whose common difference shall be = or <^ D.
Then, those numbers, which give results with contrary signs, will have
but one real value between them.. Another method of frequent appli*
OXNEBAL THEOBT OF IQVATIONS. 478
catioD; when D is a proper fraction^ -^ is to transform the equation in x
into another in y, by making x s -. Then, the differences of the yalnes
f*
of y will be greater than nnitji and only one real root will lie between
the successive integers in the transformed equation. For^ let of and
of be consecutive values of x, then aj' s= -, and a/' as ^, Hence^
T T
{pi — a/') r = y — y ^ Then the differences between the conseoatiYe
values of y is r times greater than between those of Xy and, as r is the
denominator of D; y — y must be greater than unity.
BXAMPLES.
1. Knd one irrational value in the equation, a* — 8x"+ 7a^ — 66 = 0.
Ant, 2*828.
2. Find one irrational value in the equation^ o^ + Sx* — ^ — 16a;
— 5 = 0. Am. 2-236.
8. Find one irrational value in the equation, a^-f- 2a^ — 2a:^ — 4 =: 0.
Am. 1*264.
4. Find one irrational value in the equation, a;* *— 7x + 7 «> 0.
Am. —3 048.
6. Find one irrational value in the equation, a^ + a^ — 12a:* — 17a5
— 85 = 0. Am. 4128.
6. Find one irrational value in the equation, x' — 2 = 0.
Am. 1-414.
7. Find, by the process of irrational values, one value in the equa-
tion, ac* — 4 = 0.
Ant, X =: 2, the decimal part in all the transformed equations being
zero.
8. Find one irrational value in the equation, x* + x* — 25x" — 26x
— 26 = 0. Am. 6-099.
482. If we apply the foregoing process to an example of the form,
re* + 140;* — 49 = 0, the consecutive numbers will give no change of
sign. One value in the equation of differences will be found to be
zero, and also one in the equation of the square of the differences.
40*
474 OENERAL THEOBT OF EQUATIONS.
D* and D are then both zero. When tMs is the case^ we may infer tlie
presence of equal values. On trial, as in Art. 475^ we Snd a^ — 7 ^e
greatest common divisor between the first member of the given equa-
tion and its derived polynomial. We see, by this example, that iJie
preceding method is only applicable to equations which have been freed
from their equal values.
While the foregoing method affords a complete solution to the pro-
blem of finding the irrational values of numerical equations, yet it is
of difficult application to equations of high degrees, and, in all equa-
lions, whether of low or high degrees, the difficulty increases with the
number of decimal places sought.
NEWTON'S METHOD OF APPROXIMATION.
483. This is known as the method of successive substUutwns^ and
consists in substituting, in the given equation, the natural numbers be-
tween the limits, until results with contrary signs are obtained. Let a
be the least of two consecutive numbers which give results with con-
trary signs. Then a is the first figure, or entire part of the value
sought. Substitute a + ^ for a; in the given equation, y being a small
fraction, whose second and higher powers may be neglected. Hence, y
may be found in the transformed equation, and a + y will constitute
the -first approximation to the value of x. Let a -f y be denoted by &,
and make x s & 4. y, y being a small fraction, whose higher powers
may be neglected. The transformed equation will give the value of
y, and this, with 5, will constitute the second approximation to x.
Calling 6 -f y, c, and making x as e -f y" , we can get a third approxi-
mation to the value of x, and may thus carry the approximation to as
many places of decimals as we please.
Let us take, as an illustration, the equation, a^ — 2 =b 0. Snbsti-
tuting between the limits, we find that 1 and 2 give different signs.
Then 1 is the entire part of the value. Make x = l -f y, and reject
y, we will have 1 + 2y — 2 = 0, or y = J = -6. Hence, for first
approximation, x = 1-5. Now, make x = 1*5 + y, and we will have,
•25
after rejecting y«, 2-25 -f Sy — 2 = 0, or, y « — — = 0833.
Then, for the second approximation^ we get a; = 1*5 — '0833 = 1*4167.
Pkce X = 1-4167 + y", we get
2-8834y' = — -00303889, or, y ' « — -00107.
Then, for a third approximation, x = 1-4167 — -00107 = 1*41563.
GENERAL THEOBY OF EQUATIONS. 475
484. The aocnraoy of the process evidently depends upon the un-
known qnantitj introduced being a snudl fraction. Whenever, then,
the substitution of the natural numbers make y an improper fraction,
more minute substitutions must be made, unless we intend to carry the
approximation to several places of decimals.
Thus, take the example, af + x — 8 = 0, we find that 1 and 2 give
results with oontraiy signs. Then, 1 is the entire part of the value
sought. Making a: = y -f 1, we have, y*+3y4-3y*-|-l + y +1
— 8 = 0, or, rejecting the higher powers of y, 4y — 6 = 0. Hence,
y = I = 1*5, an improper fraction. Then, for the first approximation,
we have, x as 2*5, a result obviously too great.
Next, place x =s 2*50 + 2/9 we have, after rejecting higher powers
of y, (2-50)» + 3 (2*50)V -f 2-50 -f y — 8 = 0. Then, ^ = — -512,
and, for the second approximation, x = 1-988. Making now x ^ss.
1*988 + y"} we will have, after rejecting the higher powers of y",
(1*988)» + 3 (l*988)y' 4- 1*988 + y" — 8 = 0,
or, 7*856862272 + ll-866432y" + 1*988 + y" — 8 = 0,
or, ll'856432y" = _ 1*844862272.
Hence, x/' =a — *144 nearly, and, for a third approximation, x =s 1*844,
which differs from the true value by less than one hundredth. We
see, from this example, that the error was considerable when y was an
improper fraction, but became reduced by carrying the approximation
farther. A closer approximation could have been obtained, without
canying the operation so far, by making minute substitutions. A
change of sign would have been found between the results corresponding
to 1*8 and 1*9.
LAGRANGE'S METHOD.
485. This differs from Newton's, in that the unknown expression
added to complete each successive value is fractional in form, and in
that the transformation is made into an equation involving this un-
known quantity. Thus, let a be the entire part or number next below
the value, we make x = a H — > and get a transformed equation in y,
such, that the values of y must be greater than unity. Let ( be the
entire part of the value of y, then, for the first approximation, we have
a; = a + 7 ^ — » — . Next, place y = 6 + -. The transformed
6 6 ' ^ 9 * yf
equation in y' must have its values greater than unity. Let e be the
476 aXNlBAL THEORY OF XQUATIONB.
entire part of ihe value of y. Then, approximatiYelj, y = h •{ — s
c
— — — , and, for a second approximation to x, we have, x=s a -{• - =^ a
e 1
+ -7 =■• To find a third approximate value for «, let y' = c + -^
and we may thus continue to approximate to the value of x uniO the
result is as acouzate as desired.
To apply these principles, let us take lihe equation, x* — x — 5 » 0.
We see that 1 and 2 give contrary signs, hence, re =s 1 H — . Substi-
tuting this value of x, we have 5^ — 2y* — 3y — 1 eaO. The entire
part of the value of y is 1, hence, for first approximation to x, we have
x=l + -sl4.^s2, which is plainly too great Next, making
y « 1 + 3, we get y» — 8y» — 13^ — 5 ■» 0. In this, 9 and 10
y
give contrary signs. Then, y = 1 + ^ = V> *^^ ^ ~ ^ + "A = tI>
fbr the second approximation.
Now, make y = 9 + -^^ ftnd the transformed equation in j/' will he,
4 V» — 86y» — 19y" — 1 = 0; a change of sign between the re-
sults given by 2 and 8. Hence, y = 9 J = ^j®, y = 1 + y\ = fj,
and X = 1 + ij ^ ^^y^oT the third approximate value of x. We find
on trial the third approximate value a little too great, and the second a
little too small. Hence, the true value lies between ^f and i|y and
either of these numbers differs from the true value by less than j^.
486. Let us take the simple example, a^ — 2 = 0. Place x = 1 + -»
then, y* — 2y — 1=0. We find that 2 and 8 give contraiy ogos.
Hence, for first approximation, x = 1 + } » |. Now, make y ss 2 + -^
and the equation in y will become y^ — 2i/ — 1 = 0; and, again, 2 is
the entire part of the value. Hence, y = 2 -f } ss |, and x = 1 + |
= I, for a second approximation. On trial, we find | too great, and |
too small, and these differ from each other by a tenth ; therefore, the true
value differs from either by less than a tenth. For a third approximate
value we would find, x = jj, which is a little too great, but is within
Jj^ of the true value. The fourth approximate value, j^, is too small,
but differs from the true value by less than j^^ . By continuing thus
the process, we would find the approximations alternately too great and
GENERAL THEOBT OT EQUATIONS. 477
too small^ and thus can tell at eveiy step how near we have come to the
tme values. We see that Lagrange's method enables ns to determine
the proximity to the true value at every stage of the work^ and this is
the chief advantage claimed for it over the process of Newton.
EXAMPLES.
1. of — 7a; + 7 = 0. Am, o? s= f |, after three approximations.
2. a;* — aj' — 6 = 0. -4n«. a; = J, after third approximation.
Ist approximation, 2 ; 2d approximation, | ; 3d approximation, \,
GENERAL SOLUTION OP NUMERICAL EQUATIONS.
487. When we have a numerical equation of any degree to solve, we
most first find its rational values by the process of divisors, if it is under
the proposed form. But, if it is not, we must transform it into another
equation in y, so that the coefficient of the first term shall be plus
unity, and the other coefficients entire. Then, knowing the relation be-
tween y and a;, we can determine the values of x when those of y have
been found. Next we must ascertain whether any of the values are
repeated (Art. 475). Having thus found all the rational values, we
next divide by the factors corresponding to them. Then, by means of
the equation of the squares of the differences, we can ascertain what
series of numbers to substitute in the reduced equation (Art. 481). The
final step is to find, approximatively, the irrational values by either of
the three processes explained. Now, if the number of rational and ir-
rational values be subtracted from the degree of the given equation, the
difference will be the number of imaginary values.
GENERAL EXAMPLES.
152a:'
1. Find the three values of the equation, 2aj* -= 1- 106a; ^
o
20 = 0. Am, x^\yh, and 10.
2. Solve the equation, 7? — llx* + 7a:" + 28a; — 28 -s 0.
Am. a; =. db 2, 1-856, 1-692, and —3-044.
8. Solve the equation, of + -g fa?— i = 0.
Am, « = ^> — ^7 and —8.
478 STITIIM'S THXORXM.
4. Solve the equation, ac* — 10a» — 12a" + 92x + 280 = 0.
Ans. X = 10, and 4-302.
9^s Ox*
5. Solve the equation, 2a^ — 2a^ 5- +-0- + x — l&O.
Am. x^tl, i, — }, and dtz ^2.
6. Solve the equation, a' — 4a^ — 4x» + 89aJ* — OOac* — 28a* +
2242 — 224 » 0. Ans. a; = 2, 2, and x » 2-828.
7. Solve the equation, aj* — 2a^_ 31x« + 22a? + 280 = 0.
Ans. a; = 4, and 5*134.
STURM'S THEOREM.
488. When we have, by means of the process of divisois and the
method of equal values, detected all the rational values, and then, by
the aid of the equation of differences, discovered all the irrational
values, we can determine the number of imaginary values, by subtract-
ing the number of rational and irrational values from the degree of the
equation. When this difference is zero, there are, of course, no ima^
nary values.
But this method ill circuitous, and is, moreover, rendered tedious by
the employment of tiie equation of differences. Slurtn'9 Theorem d^
iermines directly (he number 0/ imaginary values^ and dispenses with
the equation of differences.
Let X = a- + Px"-* + Q«""* + +Tx + U = 04tf an egua-
turn cleared of its equal values. Take the derivative of X, and call it
X'; then, X' = ma;--'+(m — l)Px»-"+(w — 2)Qx"-^-f -f T.
Now, divide X by X', and continue the division until we get a remain-
der in a; of a lower degree than X'. Call this remainder, with its sign
changed, X". Divide X', in like manner, by X^', and continue the
division until the remainder is a degree lower than the divisor. Chango^
in the same way, the signs of all the terms of this remainder, and csD
the resulting expression X'". Pursue the same process until we get a
remainder independent of a;, which must be so eventually, since the
given equation, by hypothesis, contains no equal values. This will be
the (m — 1)^^ remainder, and all the expressions, including X and X',
BTUBM^S THSOASH. 479
irill make up (m + 1) fanotioDS. Hence, designate by X"^', tlie last
remainder, with its sign changed. We will then have the following
expressions; Q, Q^, &c., designating quotients :
X =X'Q —X" (A),
X' ^X"Q[ —X'" (B),
X" = X'"Q" — X»^ (C),
X'" = X»^g" — X^ (D),
X»^ = X^Q»^ — X^* (E),
X--' = X-Q--' — X-+» (W).
In preparing the dividends for division, we may, as in the process
for finding the greatest common divisor, multiply by any positive con-
statu. In like manner, we may suppress any positive constant common
to all the terms of any of the successive remainders.
489. The theorem we have to demonstrate may be enunciated as
follows :
If we substitute any number, m, for x in the above series of fane-
tions, X, X', X", &c., and count the variations of signs in the resultSy
and then substitute any other number, as n, for x in the same series,
and count again the variations of signs in the results, the difference
hettoeen the number of variations in the two cases will express exactly
ike number of real values between the numbers m and n. If, then, m be
taken as the superior limit of the positive values, and n as the superior
limit of the negative values, the difference between the number of vari-
ations will express the total number of real values. When this diffe-
rence is zero, there are no real values ; when it is equal to the degree
of the equation, all the values are real.
490. The demonstration of the theorem depends upon the four fol-
lowing principles :
1. No two consecutive functions can vanish for the satne value ofx.
'EGTy if two functions, as X" and X'", could disappear together, equa-
tion (C) would give = — X>^. Hence, X»^ = 0, and X"' and X»^
being zero, equation (D) would give X'' s= 0. And so all the func-
tions in succession would become zero, until we would finally have
X"+' ^ 0. But this would indicate a common divisor, and, conse-
quently, equal values, which is contrary to the hypothesis with which
we set out
480 STURM'S THIORIM.
2. WJien mch a value u given to x <u to make any function other
than X egual to teroy the functions adjacent to the one diMappearing wiEt
he affected with contrary signs, *
For example, let X'^^ss 0. Then, equation (D) will give X"'= — X^;
and, since the own sign of X^ may be either plus or minus, ibis equa-
tion may be written X'" «= — (d= X^). Now, if X^ be affected with
the positive sign, the second member will be negative ; and, since the
signs of the two members of an equation must always be the same, X*^
will be negative, and, therefore, of a oontraiy sign to X^. But, if the
own sign of X^ be negative, then the second member will be positive.
Hence, X'" will be positive, and, therefore, of a contrary sign to X'.
8. The disappearance of any function other than X for a particular
value ofXf will neither increase nor decrease the variations of signs in
the results of the series, X, y, X", dhc.
Suppose that one of the intermediate functions, as X'^, becomes lero,
for the value x s= 5. Then, by the first principle, neither X'" nor
X*^ can be zero for x = h ; and, by the second principle, they must
have contraiy signs. There is, then, one variation of sign between X'^
and X^ when x = b. We are to show that this variation was not
caused by the disappearance of X^^. For, however little x sss b may
differ from a value of X'^' or X'^, h may be taken smaller than this
difference. And, moreover, since a quantity can only change its sign
in passing through zero or infinity, it is plain that, between x ss h and
35 s=s fc — h, neither X'" nor X^ can undergo any change of sign. They
will still be affected with contrary signs, and, for a; = 6 — h, we will
have either + X'", dz X'^ — X^ or else, — X'", db X^ + X^. We
affect X*' with the ambiguous sign, since we do not know whether its
sign is positive or negative. Now, read the double sign either way,
and take either of the preceding expressions, and you will have one
variation. Hence, before x arrives at the value 5, which causes the
disappearance of X*^, the three successive functions give one variation.
Now, when x passes beyond b, and becomes h -{- h (h being less than
the difference between h and a value of either X'", or X*), X'^ will
change its sign, but X"' and X"" will not. Hence, for a; = 5 + A, we
will have either + X'", =f X'% — X', or — X'", zp X'', + X'. And,
reading the ambiguous sign either way, in either of the above expres-
sions, we will have one variation. Hence, both before and after x
reached the value h, which made X'^ = 0, we find a variation of sign
among the three functions. Then x as 5, or the vanishing of the inter-
mediate function, X'^, has not caused that variation.
STUBM'B THSOftSU. 481
4. Every pamage o/X Arau^ zero can$e$ a ha of one variation
uihen X is CLScending towards the superior positive limits or a §ain of
one variation througho^U the series^ X, X j dfc, when x is descending
towards the s/uperior negative limiL
For^ let £c =s a be a value in tbd equafcioa, X » 0, and take u less
than the diferenoe between u and a value of X'. Theni X wiJi bave
a different sign, wb^i a + 1* is aubstituted for as, from wbat it would
have were a — u substituted for x. Henoey in tbe passage of x from
a — tt to a + u,X will undergo a change of sign, but X' will not^
since, by hypothesis, no value of X' is passed over. Neither will any
intermediate function, X", X"', &c., undergo a change of sign unless it
pass through zero, and, by the third principle, such passage ^iU not
affect the total number of variations. Therefore, in the passage of ar
from a — u to a + u, either a permanence, existing between X and
X', is changed into a variation, or else a variation is changed into a
permanence.
We will now examine the order in which this change takes place.
Designate by Xi what X becomes when x = a — u. Next, develop
an the terms of X by Art. 456; we will have Xi= A — A!u +
A'V A'"!**
— 5 — - — 5—5 + fto., (R), in wfaid A denotes what X becomes
when x = a; A' is the derivative of A, A" the derivative of A', &c.
But, since a is a value of Xy A must be equal to zero. Then, equation
A"u A"V
(B) may be written, X,= — « (A' — =—5 -f = — 5—5 — * Ac) Now,
It may be taken so small that the first term will be greater than the
algebraic sum of all the other terms within the brackets. Hence, the
sign of the quantities within the parenthesis will depend upon that of
A'. K, therefore, the own sign of A' is positive, the second member
will be negative. Then, X| must be negative, and, therefore, of a con-
trary sign to A'. But, if the own sign of A' be negative, the second
xnember will be positive. Then, X, will be positive, and X, and A'
will be affected with contrary signs. And, since Xi and A' represent
what X and X' become when a — uia substituted for x, we see that
there is a variation of signs between these functions below a value,
x = Uf which satisfies the equation, X = Q. Now, suppose x = a -j-u,
and, therefore, above a value in the equation, X = 0. Develop as
before, designating by Xg what X becomes when a -{- u takes the place
A"u A'"tt'
of X. Then, X, = « (A' + 2 — 5 + = — 5—5 + &c.), in which « is
41 2r
482 STtJRM's THEOBEM.
80 small that the sign of A' controls that of the parenthesis. We see,
that; whether the own sign of A' be positive or negative, Xt most be
affected with the same sign. Hence, when x =: a + «, and, therefore,
above a value, there is a permanence of sign between X and X'.
Therefore, in the passage of x towards the superior positive limit, a
variation is lost, and changed into a permanence between X and X',
whenever X becomes equal to zero; and, as the other functions neither
gain nor lose variations, one variation only is lost throughout the entire
series whenever X = 0.
K we had begun with a + u^ and passed down to a — «, it is en>
dent that one variation would have been gained; every time that x
reached a value in the equation, X = 0.
So, then, whenever x, increasing or decreasing by insensible degrees,
reaches a value which will satisfy the equation, X = 0, one variatioii
is lost or gained in the signs of the series, X, X', X'', &c. Hence, if
we take any number, as m, and substitute it for x in all the functions,
and count the number of variations in the signs of the results, and then
take any other number, as n, and treat it in like manner, the difference
between the number of variations in the two cases will express the
number of real values lying in the given equation between ii» ajid «.
Moreover) it is plain that if m = + cx), and n = — oo, this difference
will express the total number of real values in the equation. This
would be equally true if we used the superior positive limit and the
superior negative limit. But, there is this advantage in the employ-
ment of + 00, and — oo \ the substitution need only be made in the
leading term of the successive functions, for then the sign of this
term would control the signs of all the other terms in the same
function.
491. Having determined the number of real values, the next step is
to determine the initial figure of each one of these values. To do this,
we substitute the natural numbers, 0, 1, 2, 3, &c., until we get the
same number of variations of signs in the entire series as was given by
-|- 00. The number that gives the same variations, as + oc, is the
superior positive limit. Substitute, in like manner, 0, — 1, — 2,
— 3, &c., until we get the same number of variations as given by
— oo. We then have the superior negative limit The least of tiie
numbers, whether positive or negative, between which a variation is
lost or gained, is the initial figure, or entire part of a real value.
Should there be two or more variations lost or gained between consecu-
stuem'b thxobem. 488
numbers^ ihen there are as many yaluee as tihere are clianges of rign^
irhich have the same initial figure.
We will illustrate by a simple example.
Take equation^ o^ + Za^ — 1 ss s a;. Then its deriyatiye,
X' = 3aj* + 6aj.
^OW; multiply all the terms of X by 9, and divide by X'. After
two divisions^ we will get a remainder of a lower degree than X' ; this
ig _ 2x — 9. Change its sign, and call it X". Then, X" = 2a; + 9.
Multiply X' by 4, and, after two divisions, by X", we will have a re-
mainder, + 207. Henoe, X'" = — 207, and we will have
X » x»+ x« — 1=0
X' =3a:» + 2a;
X" = 2x + 9
X'" « — 207.
Haldng « a + od, in the leading terms of these fiinotbns, the order
of the signs will be
+ + + — ; one variation.
And for 09 = — oo, it will be
■^ + — "^j two vaiiatioiis*
Henoe^ 2 — 1 =: I9 real value.
To find the initial figure of this value, make x = 0^ 1, 2, 8, &c.,
and write the signs of the results beneath their respective functions.
"We win have
XTTf •%rrr -xrfff
then, oj = 0, — 1, 0, +9, — 207, two variations,
s 8s 1, ^• 1, +5, + 11> — 207, one variation.
Henoe, 1 is the superior limit of the value sought, and is its initial
figure. The decimal part of the value can be found by the process for
the irrational values.
The transformed equation in y^iB^ + y — 1 = 0; and that in 2,
a^ + lOz" — 1000 = 0. And, since 7 and 8 give results with con-
trary signs, 7 is the tenths of the required value. Then, x = 0*7 is
an approximate value of x.
The approximation may be carried as far as desired by the method
of irrational values.
484 BTtJBM'B THEOREM*
We need not stibstitate 0^ — 1, — 2, &c., ia the eqiutioB, riace
there is but one real value^ and that has been shown to be poative.
492. Whenever only one variation is lost between two oonsecHtiTe
numbers, Sturm's theorem affords no advantage over the prooess for
finding the irrational values, except that of detecting directlj the Bom-
ber of imaginary values. But, when there are two or more real values
comprised between two consecutive numbers, we are enabled by meaos
of the theorem to dispense with the equation of differences, whicii
otherwise must be employed to find the decimal part of those vahieB.
We will now explain the chief advantage of the theorem.
Let us take the equation, acF .— 2a^ — 2x' + ^ ^=0. We will have
the following series of functions :
X = a^ — 2x» — 2a^ + 4 *= 0,
X' = 5x* — 6x« — 4x,
X"=4a:» + 6x« — 20,
X'" = — 42x« — 168x + 300,
X' = — 2880a; + 3840,
X^ = — , a constant.
+ 00 giTes + -f- + H ) 0^6 Variation,
— 00 gives 1 h + — I fettf variationa.
Hence, there are Ai^ee real values.
Proceeding as before, we have
XT' V YW "Vt» Y»
When, a =
« x=2
+ — + — , three yariations,
+ — — -f + — ) three variations,
+ + + + — — y one variation.
Since + 2 gives the same number of variations as + <xs it is the
superior positive limit. Moreover, as there are two variations lost bc^
tween 1 and 2, there are two real values between them. Had we sub-
stituted the natural numbers, 0, 1, 2, 3, &c., in the given equation, we
would have found no change of sign, because an even number of values
lay between consecutive numbers. To detect these values, without
the aid of Sturm's theorem, we must either have recourse to the equa-
tion of differences, or to minute and tedious substitutions.
\
\
Soy^ then, ii is evident that v)henever an equation ooiUaim twOyftmr^
«tJK; or any number of even values between consecutive numbers, Sturm's
theorem enables us to detect Aese values in the shortest poss^k manner,
493. But; in addition to tliis, tbe theorem gives us the means o(
finding the decimal part of the irrational values in a shorter and better
manner than by the usual process^ as we will now show.
Since 1 is the entire part of both the irrational values, the first step,
according to the usual process for finding the irrational values, is to
transform the equation in x into another in y, so that the values of y
shall be less by unity than those of x.
The transformed equation in y is
y* + 5y* + 8y > + 2y« — 5y + 1 =, 0,
and that in z, is
^8 + 50^* + 800a» + 2000;^ — 6OOOO2 + 100000 = 0.
We find, on trial, that 0, 1, and 2 ^ve positive results; 8 and 4
give native results ; 5, and all numbers above 5, give, again, positive
results. Hence, 2 and 4 are the tenths of the sought values; and
1-2, and 1*4 are those values, approzimatively.
Now, if, in the transformed equation in z, two values had lain be-
tween consecutive numbers, we must have had recourse either to
minute substitutions, or to the equation of differences. If, for instance,
the second value had been 1-26 instead of 1*4^ the tenths would have
been the same for both values^ and the substitution of the natural num-
bers in z would have given no change of sign.
To obviate a difficulty that might ocour in more ihan one of the
transformed equations, we proceed thus,
We take all the fanctions, X, X', X", &c., and transform ihem into
others in y, so that the values of y shall differ from those of x, by th^
initial figure, which, in this case, is unity. We will then have the
series,
Y«y*+5y* + 8y»-f-2y»^5y4-l,
T' = 5y* + 20y»+ 24y> + 4y — 6,
Y'' = 4y'4-18y" + 24y_10,
Y'" = — 42y« — 252y + 90,
Y'- = — 2880y + 960,
Y' 5= — ^, a constant.
41*
.^
486
stcem's thbobxm.
1
3 may
After liaying found Y^ as indicated^ we inaj get its deriTatiye, Y*,
and then divide Y by Y'^ and prooeed as we did when getting the
fiEmotionS; X, X'^ X'', &o, Bui, in general, the better method is to
txansform the functions in x into others in y, so that the values of jr
shall be less than those of x, by the initial figure of the sought value.
The transformed fanctions in 2 are :
Z = ;?• + 50z* + 800z» + 2000z«-
- 50000z + 100000,
Z' = 5«* + 200z> + 24002» + 4000z — 50000,
Z" = 42» + 180z» + 2400z — 10000,
Z'" = — 422» — 2520z + 9000,
Z^'s— 2880Z + 9600,
Z* =. — , a constant.
And we have the following results :
Z, Z', Z", Z'" Z",
Z\
When z =
+ — — + +
— , three vaiiations,
« « = 1
+ — — + +
_ (( it
« 2 = 2
+ — — + +
c< u
" z = B
— + +
— , two variationg,
a z = i
+ _ _
_ u a
" z = b
+ + +
— f one variation.
We need go no further in the substitution, since 5 gives the same
number of variations as + <>o« We see that there is a variation lost
between 2 and 3, and another between 4 and 5. Hence, 2 and 4 are
the tenths in the required values, as we before found. Now, if there
had been two, or any number of values having the same initial decinuJ
figure, 8, for example, there would have been as many variations lost
or gained as there were values having the same initial decimal.
494. We see, then, the two great advantages of Sturm's theorem in
finding the irrational values over the other three processes described :
1st. When an equation comprises an even number of values between
two oonseoutive numbers, it enables us to detect these values without
minute substitutions, or the employment of the equation of differences.
2d. When two or more values have the same initial decimal figure, it
enables us to tell the exact number of those values. If we add to these
two advantages the one first mentioned, that of detecting directly the
STUBM'B THSORSil.
487
nnmber of real, and, conseqnend j, the number of imaginary yahies, we
can see how important the theorem is.
To get the hundredths, the fdnctions in z may be transformed into
others in «, so that the yaines of a shall differ from those of z by the
initial decimal figures ; in this case, 2 and 4. Then, again, transform
the functions in s into others in to, so that the values of to shall be ten
times greater than those of s. We will then have a series of functions,
W, W, W", W", &c., in which we may substitute the natural num-
bers, 0, 1, 2, &o., until we get as many variations as + oo gives in the
series. The least of the consecutive numbers, between which a loss of
Tariation occurs, will be hundredths in the sought value. We, of
course, will have two series of functions in w, the one dorrespondlDg^to
2, as the initial figure of decimals, and the other to 4.
495. Since there were three real values in the equation, sc" — 2x'^
226* 4- 4 ss 0, and we have found but two positive values, the other
must be negative. To determine this negative value, let us resume
the functions,
X - «• — 2a» — 2x« + 4 - 0,
X' = 5x* — 6a:« — 4a;,
X"=:4a:» + 6x« — 20,
X'" ^—A2a^ — 168» + 800,
X'^ « — 2880a; + 8840,
X* ^ — , a constant.
Substituting 0, — 1, — 2, — 3, &o., in the foregcdng series, we will
have
XY' TT" Y"' Y»» T^
When, x=
« x = —l
« «=— 2
+ — + + — , three variations,
+ + — + + — , three variations,
— + — + + — , four vaiiatioos.
And, since — 2 gives the same number of variations as — oo, it is
the superior negative limit. And, since a gain of variation occurs be-
tween — 1 and — 2, the entire part of the negative value b — 1. In
this case, we know that there is but one negative value ; we may then
ptt)ceed at once to find the decimal part of this value by the usual
process for irrational Values, without forming the functions, Y, Y',
Y", Ac.
48S 8«9RK'» TBBOmiM.
The tnmsjtbnned equatoi in y is
y» — 5y* + 8y — 6y» + 8y + 8 = 0,
and tliai in z,
/ _ 50z< + 800z» — 6000«» + SOOOOz + 300000 = 0.
A change occors between 4 and 5. Hence, 4 is the tenths of tiie
sought yalne, and we hare x = — 1'4 for the approximate vahie.
Now, diminish the values in the equation in 2r by 4, and the tzans-
fbrmed equation in $ will be
. ^ — 20^ + n60«»~ 21040^ + 180480j-f2168«=;0,
and that in to will be,
fc»— 200w'+176000w»— 21040000«>H130480O000tt?+21536000OO:sO.
A change of sign occnis between the results after the substitution
of — 1 and — 2. Hence, 1 is the hundredths of the negative vahie.
And we have, for a second approximation, x =3 — 1*41.
Remark,
496. There is one point of considerable importance in the demon-
stration, which deserves to be attended to. It has been shown that a
variation is lost or gained between X and X', every time that X be-
oomea equal to aero, or that x passes a value. It might be aaked, tiken^
Why not confine the substitutions to X and X' ?
It is to be observed,, that a change of variation only takes place be-
tween X and X'' when very minute substitutions are made. For, tn
the dem(«stration of the fourth principle, we supposed ii to be indefi-
nitely small. Now, if we substitute a number, p, which will make the
signs of X and X' contrary, and again substitute another, p', there being
two values of X, and none of X', between p and p', then X and X' will
still be affected with contrary signs. So, that, between p and j/^ there
will be no change of variation, though there are two real values. But^
if one value of X' lies between p and p', there will be one variation
lost, and but one. For, whilst X changes its sign twice, X' will change
its sign once.
Take X = 6a;»_5jr + 1; then, X' « 12* — 5.
Tbore ia a yaiintion be^oon X and X^ wheD « ■■ 0, and lUl is
changed into a pennanence when x 9b 1. Iheie are two values e( X
between and 1^ and one of X\
Again, if there were three, five, or any odd number of values of X
passed over, and none of X', there would \» one variation lost Of
gained, and but one. But, if, at the same time, an odd number of
values of X' were passed over, there would be no change of variation.
It is plain, then, that the loss and gain of variation between X and X'
will only correspond to the number of real values, when the aiibstita*
tions are so minute that no value of either X or X' is contained be-
tween them.
Sch^wm.
497. Whenever any function is constantly positive for all values of
X, we need not form any other functions, but only count the number
of variations given by + oo, and — oo, in the constantly positive fane-
tion, and in the functions which precede it. For, if we formed the
succeeding functions, they would give the same number of variations
for + 00 that they would for -^ oo. Hence, the differenee between
the number of variations of these functions must always be zero. To
show this, we take for granted that a function, which always remains
positive for all values of x, must be of an even degree, since it will
contain only imaginary values. The next succeeding function may
have its leading term either positive or negative, but the next must be
n^ative, since any value that reduces the function, consecutive with the
positive one to zero, must cause the adjacent functions to be affected
with contrary signs. All the functions of an odd degree may be either
positive or negative, but those of an even degree must be alternately
positive and native.
Let of + nu^ + &c., be the constantly positive function. We wiO
then have the series,
+ a^ + flias^ + *e. a. X%
±n3* + kc — X'+»,
— px*— 4c. — X'^,
+ «:" + 4c. ^ X'^,
±to +4c. — X-^,
— A, a constant, ^ X*^,
492 BlSSGABTia' BVLB.
When a term k missing from an equation, it must be sapplM irith
the coefficient, d: 0. Tben, in reading the signs, we must first count
the variations and permanences, regarding the ambignoos sign as posi-
tive, and then count again, regarding the ambigaons sign as minus.
Thus, taking the equation, a:" — 4 = 0, we must write it of±Ox —
4 aas 0. Counting the upper sign, we have a permanence between the
first and second terms, and a variation between the second and third
terms. Counting the lower sign, we have a variation between the first
and second terms, and a permanence between the second and third
terms. In whatever way, then, we read the ambiguoiiA sign, ibev« will
be one permanence, and one variation.
If there are any number of missing terms, they must be supplied in
like manner with positive or negative zero coefficients.
499. To demonstrate the rule of Descartes, it is necessary to show
that the multiplication of any equation by a factor, corresponding to a
negative value, will introduce into the new equation at least one more
permanence than existed in the old ; and that the multiplication by a
fector, corresponding to a poi^tive value, will introduce at lecut one
more variation.
1st Let us take the equation, a;" — Px»~' + Qa;»~" + ILc*"* —
Sx»^ . . . . + Tx — U = 0, and multiply by a factor, a + a, corres-
ponding, to a negative value. The resulting equation will be
+a
—Pa
+Qa
X--*— S
+Ea
X--* +Tbi^— U|x=0
— Sal +Tal— Ua ^^
Now, it is plain that, if we suppose the coefficients in the upper
column are, throughout, greater than the corresponding coefficients in
the lower column, there will be the same number of permanences in
the new as in the old equation, until we get to the last two coefficients
(the coefficients of x and x°), we will then have one more permanence
than in the given equation. And, if we suppose the coeflicientB of tbe
lower column to be greater throughout, there will be a new permanenoe
between the first and second terms, and the same suooession of signa in
the remaining terms. Moreover, it is evident, that if the lower ookmiB
sometimes prevailed, and sometimes did not, there might be more than
one permanence introduced. For instance, if a were greater than P,
Pa less than Q, Ba greater than S, T greater than Sa, Ta greater than
IT, there would be but one variation in the new equation.
DXSCABTXS' BULJi. 40i
It is even possible to cliange all the ^'aiiations into permane&ces, by
xnultiplicatioii by a factw oortesponding to a negative value. As an
illustration, take the equation, af — 2x* + 12x* — Stc* 4- 36a:^ — cc +
15 s= 0, and multiply by x + 4. The new equation will be,
x'' — 2 x« + 12 a*— 8 a* + 36 a»— 1 x« + 15 a; =
+ 4 _ 8 +48 —32 +144 —4+60'
or, x* + 2a^ + 4a^ + 40a^ + 4a^+ 148a* + 11» + 60 = 0.
By this example wo see that an equation, containing only variations,
is changed into another containing only permanences, by multiplying
the former by a factor corresponding to a negative value. Seven per-
manences have been introduced where none existed before. And, by
recurrence to equation (M), we see that it is impossible to read the
signs in any order, without having one more permanence than in the
given equation. Now, the given equation may have been one of the
first degree, m being equal to one, and P, Q, R, S, and T being equal
to zero. Then, if the sign of U were positive^ there would be one
n^ative value, and conversely. The new equation (after multiplica-
tion by X + a) would be of the second degree, and would contain, at
least, one more permanence than the pld. And, by multiplying this
new equation by another factor corresponding to a negative value, we
would introduce, at least, one more permanence. And so, by continu
ing the process, it could be shown that, whatever might be the degree
of the equation, the number of permanences must always be equal to,
or exceed the number of negative values.
500. 2d. By a similar course of reasoning, we cotxld show that ih^
the multiplication by a factor corresponding to a positive value would
introduce, at least, one variation ; or, in other words, that the number
of positive vahies can never exceed the number of variations.
Take, as an illustration, the equation,
x« — 4a;* + 12x* + 8x» + 30x« — a; + 15 = 0,
and, multiply it by the &etor, 26 — 2, the resulting equation will be,
x^ — ex^ + 20x» — 16x* + 14a;* — 6W + 17a;— 30 = 0,
and has gained three variations.
501. It is plain that the preceding reasoning has been on the sup*
position that a was a real value, otherwise we could not, in equation
42
494 DCSCABTSS' RULE.
(M)y have institated any comparisonB between a and — P, — Pa, and
+ Q, + Ha, and — S, &e. In cajse, then, that there are imaginaij
values in an equation, the rule of Descartes only points out limits, be-
yond which the positive and negative values cannot go. But, when
the equation contains only real values, the number of positive values
will be exactly equal to the number of variations, and the number of
negative values exactly equal to the number of permanences. To show
this, let m = degree of the equaiion, n = number of real negative
values, p = number of real positive values, h = number of vaiiationB,
h' ss number of permanences. A slight inspection will show that, in
case of real values, b + h' s= m; and we know that n + j? = m.
Hence, b + h' = n -{-p. But, since n cannot exceed I/^ andp cannot
exceed b, we must have n = y. For, if n <^ y, then necessarily
p^by which cannot be. In like manner, we could show that we
must hAvep = &.
CaroUary.
502. In case of there being one or more missing terms in an equa-
tion, the rule of Descartes will enable us to detect imaginary values.
We have only to supply the missing terms with plus or minus lero
coefficients, count the variations and permanences when the upper sign
is taken, and then again, when the lower sign is taken. If there be
any discrepancy in the results, there will be ima^naiy values. Thus,
take 0?* + 4 = ; supplying the missing term, we have a^ ik Ox -f 4
r= 0. The upper sign taken in connection with the other two, gives
two permanences, whilst the lower gives two variations. These dis-
-orepant results indicate imaginaiy values.
Take aj* + Bos" — 4 = 0, then, a* db Oo^ + 8x" d= 0« — 4 = 0.
In one case, we have one variation and three permanences; in the
other, three variations and one permanence. The difference in the two
readings again indicates imaginaiy values.
GENERAL EXAMPLES.
1. What are the values in the equation, x" — 2a? — 7a; + 1 = 7
2. What are the values in the equation, x* — 7a; + 1 = ?
3. What are the values in the equation, x' — M + 7af — 8aJ* +
9a;» + 10a;" — 11a; + 12 = 0?
ILIMINATION BITWXXN TWO XQUATI0N8. 495
4. What are the YBlues in the equation^ nf — 1 = 07
6. What are the values in the equation, x* — 1 = 07
6. What are the values in the equation, a:^ + 1 = 7
7. What are the values in the equation, a^ + 4a:' + 6a^ + 4x +
lr=07
ELIMINATION BETWEEN TWO EQUATIONS
OF ANY DEGREE.
503. When one quantity depends upon another for its value, it is
said to be a function of the quantity upon which it depends. Thus, in
the equation, y =i 2a; — ^4, y is a function of a:, because eveiy change
in the value of x will produce a corresponding change in the value
ofy.
The mathematical symbol to designate a function may be F, or/^
or the Oreek letter, t* Thus, to indicate that y is a function of a;, we
may employ the notation, y = F(a;), or y =f(x)j or y =s t(2j). The
second member may contain consianUf as well as the variable, x. Thus,
^ is a function of a; in the foregoing equation, y as 2a; — 4, the con-
stants being 2 and — 4.
504. TUe most general form of an equation of the m^^ degree be-
tween two variables, x and y, is,
a- -f Bx--* + Cx--« + Dx--» + Ex--^ + U = 0;
in which B, C, D, &o., are functions of y.
B is supposed to be of the first degree in y, and of the form,
a + hy.
is of the second degree in y, and of the form, c + dy + e^.
D is of the third degree in y, and of the form, /+ gy + Ay" + /y*.
E is of the fourth degree in y, &c. &c.
U is of the t»** degree in y, and of the form, u + my + +y">
and it does not contain x,
505. The equation is said to be complete when x enters into all the
terms but the last, y into all the terms but the first, and when, also,
the sum of the exponents of x and y in each term is equal to m.
486 ELIMINATION BETlfXIII TWO XQUATI0N8.
506. Elimination between equations of a degree faigker than the
first \b nsually effected by means of the greatest common divisor. This
method of elimination has already been explained (Art. 214), but we
propose to demonstrate the process more rigoronslyi in two wayB..
1. Let A as 0, and B = 0, be the proposed equations containing
both X and y.
Now, if we knew beforehand a value m of y, that was common to
the two equations, A &a 0, and B = 0, and substituted this value in
them, the new equations, A' = 0, and B' = 0, would contain only x
and constants. Now, if n be a value of x, in the equation, A' := 0,
its first member must be divisible by x — ^ n, and it is plain that the
given equations would not be simultaneous unless n would also satisfy
the equation, B' = (Art. 207). Hence, x — n must also be a divi-
sor of the equation, B' = 0. We see, then, that the hypothesis of a
common value in y results in the condition of a common divisor in x.
Conversely, if we can force the given equations to have a common
divisor in x, they must have a common value in ^. It is upon this
principle that we seek for a common divisor between the first membexs
of the equations, A = 0, and B s= 0, and continue the process antil
We get a remainder freed from x. It is plun that, if we place this
remainder equal to zero, and substitute the value of y, found from it in
the last divisor, it will be an exact divisor, and will be the one sough!
From the foregoing reasoning it is evident, that if it be absurd to
place the remainder, freed from x, equal to zero, the given equations
are not simultaneous.
507. 2d. Let the successive quotients, in the process of dividing A
by B, B by the remainder, &c., be designated by Q, Q^, &c., and the
successive remainders by R, R', &o. Then we will have.
A « BQ + B, (M)
B « Rg + R', (N)
R^R'Q^' + R", (0)
&c. &c.
Now, since, by hypothesis, A = 0, and B =s 0, equation (M) will
give R = 0. And, since B « 0, and R = 0, equation (N) will give
R' ss 0. From this we see that we have a right to equate, with lem^
that remainder which is freed from x, and contains only y.
508. The above series of equations show, moreover, that if the x^
mainder, which is freed from x, and the preceding divisor be piMed
or ANT PBQBSE. 497
eqtial to zero, tihe system of values so found will satisfy the ^ven equations.
For, if B." be that remaindery and R' the preceding diyisor, when Bf'j
and Bf are equated with sero^ equation (0) shows that B also = 0.
And B and R^ being equal to zero, from (N) we get^ B ss 0. And;
since B = 0; and B = 0^ equation (M) shows that we will also have
A = 0.
Hence, the values of y, found by placing the last remainder equal to
seroy may be substituted in the preceding divisor, in order to deduce
the corresponding values of x. The importance of this remark consists
in the fiict, that the preceding divisor is of a lower degree than either
of the original equations, and, therefore, more readily solved.
509. The reasoning in Art. 507 proceeds upon the supposition that
the successive quotients, Q, Q', Q^', are all finite, and then, of course,
the successive products, BQ, B(/, B'Q'', &c., will all be zero, when
B = 0, B = 0, R' = 0, &c. But, if any of these quotients be frac-
tional in form, it may happen that the value of y, found from placing
the last remainder equal to zero, will reduce the denominator of the
preceding divisor to zero also. In that case, we would have the
product of zero by infinity, which is indeterminate.* Suppose, for
4
example, B = y — 1> Q — , . Then, A « BQ + E becomes
A = B(-^-^) + y — 1; or, whenBs=0, A=xB^ = B oo; or
(since A and B are zero), == oo, which may, or may not be, a true
equation.
510. To avoid the fractional form of quotient, it may sometimes be
necessary to multiply the dividend by some function of y, though this
multiplication may possibly introduce some foreign values into the
equation, as will be shown more fully hereafter.
511. We will now illustrate the foregoing principles by a few
examples.
* Let OD = Ay diyiding both members by 0, we get oo = -^ = oo, a true
equation. And this will evidently be tme when A = 1, 6» 10> 20, or anythmg
whatever.
42 * 2o
49S XLIHINATIOK BITWXSN TWO EQUATIONS
Let a:' + 5^ — 8 — 0, and 2je — 8^ + 2=:= 0.
4
4aj«+V— 82 2a: — 8y + 2
4ae» +6xy+4a; 2a; + (3 y — 2) = Qaotient.
Igt Remainder «« 2(3y— 2)a;+4/— 32
2(3y-2)x~V+12y^
2d Remainder » 13^— 12y— 28=0.
From wbicli we get y = 2, and y b — 1| ; and these, when sub-
stituted in either of the given equations, give xf ^2^ and 2/' ss — a|.
Let x" — y* — 7 = 0, and as — y — l — ssO. Combining, we
will get as" + (y + l)x + (y* + J/) + (y + 1) for a quotient, and
y<^ y .— 2 for a remainder.
The equation, formed by placing the remainder, freed from Xy tqutl
to zero, is called the ^na^ equation. Jn this example, the final eqa^
tion, ^ +y — 2 =s 0, gives ^ as 1, y" = — 2. And these vafaus
for y, when substituted, give a/ = 2, and a/' «« — 1.
Let ac» — 8yx« + 3yx — 5x" + lOyaj + 6aj— y«— 5y« — €y = 0,
and af^^bya? + 8y*a — x — 4y* + y a= 0. Then the first qaotienl
is + 1, and first remainder (2y — 5)ac* — bj^x + 7x + lOyx + 3^
— 5^ — 7y . Preparing the last divisor for division by multiplying by
(2y — 5)*, we have
(2r-6)*
and this, when divided by (2y — 5)x" — 5y*a; + 7a; + lOyx + Sy
— 5^, gives as a quotient, (2y — 6)x — 5y* + 15y — 7, and as a ie>
mainder y^x — lDy*« + 35y*a; — 50ya; + 24x — y* + lOy* — 35y' +
50yB — 24y. And, by factoring this remainder, we get (y^ — lOy* +
85y« — 50y + 24)x —yit/" — 10y» + 35/ — 50y + 24) ; or, («— y)
(y* — lOy* + 35y» — 50y + 24). Rejecting the factor, x — y, as
leading to arbitrary values, we have the final equation, y* — • lOy* -f
Zb^ — 50y + 24 = 0. This equation, when solved by the prooesi
of divisors, gives / = 1, y" = 2, y"' « 3, and y>^ «= 4. And these,
when substituted, give a/ = 3, a/' = 5, of" = 5, x*^ = 7.
612. Two things are suggested by this example. 1st. May not tibe
multiplication by the factor, (2y — 5)^, involving one of the unknown
quantities, have introduced foreign values ; that is, values which did
not enter the given equations ? 2d. x — y being a common fiustor to
or ANY ]>B0BS1. 490
the renudndor; is foaad to be abo ooounon to both die giyen equatioiHi.
How are such fiiotora to be treated ? We will examine these sabjeets
separately.
513. Ist. In regard to foreign values^ we have this simple test
Place the multiplier equal to zero, find the value for y, and, firom either
of the given equations, the corresponding values of a; ; if these values
be the same as some of those found from the final equation for y, with
the corresponding values of x, then the system of common values in
both X and y must be rejected. In the example, placing (2y — 5)'
ss 0, we get y a I, a value dijSerent from those before found. HenoOi
the multiplier has not introduced a foreign valuer But, if the multi-
plier, placed equal to zero, had given us, for example, ^ = 1, with the
coiresponding x = 3, then this system of values must be rejected.
514. How are &ctors to be treated which are common to both of the
first members of the given equations ? There may be three cases, but
all lead to arbitrary values. 1st. The common factor may be a func-
tion of X only, /(x). 2d. It may be a function of y only, X^)' ^d.
Jt may be a function of both x and y, f(x, y).
When the common fiictor is f{x) only, x will have determinate
values, and y indeterminate. For, by placing /f x) = 0, we will get
true values for x; but, wheny(x) = 0, both equations will be satisfied,
-whatever values y may have. Take the equations, (a + hx) (a^ -f
2^05 — ay) = 0, and (a + bx) (4a:y — 2yx + my^ « 0. Placing
a + &e ss 0, we get x bs -.. — , and both equations will be satisfied
when a + 5x ss 0, whatever may be the values ofy. Hence, /(x^s^O
ffVBB X determinate, and y indeterminate. In like manner, a common
fiietor, /(y); would give y determinate, and x indeterminate. In the
ihe third case, the common fskctor, f(Xj y) a 0, will satisfy both equa-
tions ; but| since we have a single equation, f(x, y) b 0^ oontaiBing
two unknown quantities, the values of both x and y must be indeter-
xninate. Thus, take the equations, (2x — 4y) (x* + ly — 5) = 0,
and (2x — 4y) (xy — Tx" + 5x* — 7y") = 0. It is plain that both
equations will be satisfied when 2x — 4y ss ; bpt the equation,
2x — 4y = 0, will give indeterminate values for both x and y.
Hence, we conclude that, when the given equations contain a com-
mon factor, it must be divided out. For, a common factor, /(x), would
give determinate values for x, but indeterminate for y; a common
factor, y(y); would give determinate values for y, and indeteimiAate
600 ELIMINATION BETWEEN TWO EQUATIONS
toT x; and a common factor, /{^X; y); would give x and y, boih inde-
terminate.
515. Take the equations, (x — V) Qx? — 2xy + x — 2) s 0, and
(x*'^— ay — 2) (x — 1) = 0. Suppressing x — 1, we have x* — 2xy
+ X — 2 = 0, and x* — xy — 2 =: 0, fix)m which we get y » 1, and
« = 2.
Take the equations, (2y — 6) (xy + 5ary — y -f- 1) = 0, and
(2y — 6) (xy + 4y — 4) = 0. Suppressing 2y — 6, and combining
the resulting equations, we get 16^" — 53y + 87 = for the final
equation. From which, y = 1, y" = | J, a? = 0, x" =« — f^.
Take the equations, (2x — 7y) (xy — y + 5x* — 5x) = 0, and
(2x — 7y) (ay + 7x — 7 — y) = 0. Bemoving the common factor,
we get, after combination, Sx* — 12 + 7 = for the final equation.
From which, a/ = ^, x" = 1, and, by substitution, j/ = — 7, and
516. When the first members of the ^ven equations can be resolved
into factors of the first degree, or of a low degree, the elimination will
be greatly facilitated.
Take the equations, (x — 1) f yx — 8) (x" — 2xy) =» 0, and (yx —
2x) (x« — 2y) = 0.
These equations* can, obviously, be satisfied when
X— 1=0
andyx — 2xss0
when X — 1 =
and x' — 2y =
when yx — 3 =b
(A)"^ ^_«:._«(B).„^ f o,_n(C).
and yx — 2x8bO
whenyx — 8=0 ^. whenx* — ^2xy=0 -, when x" — 2xy=0I — .
and x»— 2y=0 ^ -^ and x«— 2y=0 ^^ and yx— 2x=0r^
From (A) we get the system of values, x = l, and y = 2.
From (B) « « « x = 1, and y = }.
•From(0) " « « x = }, andy = 2.
From(D) " « « x=t/B, and x = -4=.
From(E) « " " x=:l, andy = J.
From(F) « " " a^ = 0, x" = 4, andy = J,
y" = 2.
An artifice will sometimes enable us to decompose the first memben
of the given equations into their respective factors.
Of AMT DXQBXX.
501
Take the eqnfttion, 7? — 2ya5 + 6a5 + y* — fljy + 6 = 0, (A), and
of-^-^yx + y" + 6a? + 6y + 5 = 0, (B). From (A), we get as"— 2yaj
+ y* + ^ — €iy + 5 = 0, or (x — y)' + 9(« — y) + 9 — 4 « 0;
or (since the first three terms oonstitate a perfect square), {x — ^+8)'
_4 = (x— y + 3)«— (2)* = (a?T-y + 5)(x— y + l)(Art. 60).
From (B), we get (x + y)* + 6(x + y) + 9 — 4 = 0, or (x+y + 3)"
— 4=:(x-f yH-Sy— (2)"=(x+y+5) (x+y+1) (Art. 50). Hence,
we have (x — y + 5) (x — y + 1) = 0, (A'), and (x + y + 5)
(x + y + 1) = 0, (B'). And (A') and (BO will evidently be satisfied.
(M)
when X — y+5«s0
and x+y+5sB0
and when x — y -f 1 r=
and X + y + 1 s
when X — y+5j
and x+y+li
(N)
when X— y+l==0
and x+y+5»0
(P),
From (M) we get the system of values, x =
From (N) " « « x »
From(P) " « « xi
Fiom(Q) « « w xss
5, andy
8, andy
8, andy
1, andy
0.
+ 2.
— 2.
0.
Take the equations, a? — 8yx — 8x + 2y» + 7y — 4 = 0, (A),
and Q? — 4x + xy — 4y = 0, (B).
From (A) we get (by making 2y* == -^ ^, and adding and sub-
Ixaoting j)f
.(■
or, IX
2
i)-(f
5*
)=<a?-^f— 4) (x— 2y+l) (Art. 50).
From (B), we get x(x — 4) + y (x — 4) = (x + y) («
Hence, we have (x — y — 4) (x — 2y + 1) = 0.
and (x + y)(x — 4) = 0.
From which we get the equations.
-4).
fb— y— 4=0
and a!+y=0
»— 2y + l =
»— 4 =
_, as— y-^4 =
W. «_4=0
(S),
«— 2y+l=0
a;+y =
(T),
(U).
602
XLIMINATIOn BBTWBSN TWO EQUATIONS
From (B)| we get the system of YJftlaeS; x =3 2, mnd y = — 2«
From(B), « " " «=:4, and y = 0.
From(T), " " " a = _^, andy = + 1.
From(tJ), " " " a; = 4, andy = |.
Take the equations; ^ + y*.*-2yx — 4y + 4x = 0, (A),
and ay — t/'~6ajy + 5 + 4y = 0, (B).
Prom (A), we get (% — yf + 4(a; — y) =0,
(«-y)(«-y+4) = 0,(A').
And from (B); we get (by adding and subtracting 4),
«y — 6xy + 9 + 4y— y — 4=0,
or, (ajy— 8)«— (y— 2}*=0, or, (^+y— 6) (rry— y— 1)=0, (F),
(AO and (B') give the system o£ equations,
and a?y + y — 5 =
./IN ^— y=^^Tn aj — y+4=0
^ '''a:y-y-l=0^^^'«y+y — 5=0
Wi
flc— y + 4 =
ay — y — 1 =
(K).
From (G), we get the values,
,a^ =
^==l+^,y.^
— 1 — s/21
2 '
— 1 — v'2r
From (H)| we get the values,
a^ =
,a/' =
y=4^,^'
From (E), we get the values.
a/==U^,>
1— v/6
1 — V'S
y=+
2
8+s/29
,*"= +
— 5 — ^/29
2 *
8 — v'W
OV AKT DXOBBS. 608
517. One of the equations only maj be capable of decomposition
into&ctoTS.
The equations may be of the fonn^ A = 0, and BD = 0. We will
then have two systems of values; one resulting fiom A = and
B = 0, the other from A = and Ds=;0,
Take ihe equationii^
aV— 7rry — 20a; = 0, (A), and (ay — 5a5+8) (a — 2) = 0, (BD).
We get the system of equations^
a^y _ 7xy — 20x = 0, (A), and ay — 6a; + 3 =0, (B).
From which, 0^ = 8, and*" = |, ^=^4, mi^^s — |.
(A) and (D) give xy — 7a:y — 20x = 0, and a; — 2 = 0.
From which we get a; = 2; y' = — -^ , y^= j .
618. If any of the successive remainders be capable of decompo-
sition into factors, which are functions of x or y, these factors may be
placed, separately, equal to zero, and the deduced values of x or y sub-
stituted in the preceding divisor.
For, let B be one of the dividends, R' the divisor, Q the quotient,
and/(x) X y(y) the remainder.
Then, |, =Q +-^f^||^, or R = K'Q +/(x) x/(y).
And this equation will be satisfied when B' = and/^x) a 0, or
when R' = and/(y) = 0.
Taketiie equations, yx* -#-yW — as" — y^x + yac + y* — 1=0, (B).
and aj» — y + l=0, (BO.
IHviding R by R', we will get a remainder, (cc" + 1) (y* — 1)
Placing «■ + 1 = 0, we get a5 = db ^ — 1, and this, substituted in
(R), ^ves y = 0. Placing y* — 1 =z 0, we get y ■= db 1, and these
values, when substituted, give a/ = 0, and V = db ^ — 2.
Hence, we have the system of values.
a/=+^— 1
y' =
a/'=— v'— 1
y^ =
a/" =
y'" = l
x»^ = ^/— 2
X^=:— v/— 2
y»^== — 1 y'=— 1
All of which will satisfy the given equations.
504 XLIHINATION BBTWEBN TWO BQUATIOKB
519. Wben it is necessaiy to mnltiply (A) by dther a functioii of
X or y to make it divisible by (E), we can tdl whether the mnltipfier
has introduced foreign values, by combining it^ placed equal to lero,
with B = 0. If any of the values thus found are the same as those
resulting from the combination of the given equations^ we must xejeet
these common values from the solutions of the ^ven equations.
For, let A = 0, and B = 0; and suppose that (A) will not be divi-
sible by B until it has been multiplied hj/(x). Then we will have
A(/c) = 0, and B = 0, which can be satisfied when A=0, and B bO;
or, when /(x) = and B = 0.
Now, it is plain that, if the combination of the new equatioiii
A/(x) = 0, with B == 0, gives, among its system of values, the same
as given by /(a;) = 0, and B = 0, we must reject the common values
as having been introduced by the multiplication of /(x).
Take the equations, x'y — 2xy + a:* = 0, (A), and a^y -^ 2a^ +
X = 0, (B).
Multiplying (A) by x", to prepare tor division, we get for the final
equation, a^(x* H- 2x" — 5jc + 2) = 0. From which we get, x = 0,
ixf = l,af' = ~^'^'^^'^ ,af" = ~^~^^^ . The oomspondiiig
values of yare,i/ = fet/-=^^"-^^t/-- = -(^^+^^^
^ '^ ^'^ 38 — 10^17'^ SS + lOvOT
(M).
Now, placing /(x) = 0, and B = ; or, x* sas 0, and a^ — 2j^ +
X = 0, we get X = 0, and y = } ; ^^^f Binoe these values are the same
as the first of those, marked (M), we must reject them from (M), and
leave but three values for x, and three for y,
520. If A be exactly divisible by B, the values of x and y will be
indeterminate.
For, then, ^ =Q; and, since A=:0, and B = 0, we will have
J = Q, or = OQ . (P). Now, it is plain that Q may be /(x), or
J{y)f ory(x, y). But, equation (P) will be satisfied, whatever may
be the form of Q, and whatever may be the values of x or y.
Take the equations,
x* — 8x"y — 2x" + 8y»x» + 6x«y — y'x — 6y«x + 2y» = 0,
and x" — 2xy + y* = 0.
We get as a quotient, x" — xy — 2x + 2y, and a remainder aero,
and any value whatever of x, with the corresponding or deduced value
of y, will satisfy both equations.
09 AMT BXa&EB. SOS
Remark,
The caae ezliibited in 520, differs only from that in 614 in this
respeoty there is a oommon &ctor to ihe two equations in both instanoesi
but 520 does not manifest that feotor.
The indeterminate nature of ihe given equations, when the final
equation in y is zero in both members, may also be shown by retaining
ihe trace of y. For, when we place the zero remainder equal to ierO|
we haye Oy = 0, or y = ^.
521. When the final equation reduces to a constant, the value of y
wiD be infinite, and the given equations will be contradietoiy.
For, then we will have Oy = A, constant; or y = oo, the symbol of
absurdity.
Hence, the combination of the given equations has led to an abeord-
ity, and, therefore, these equations must be contradietoiy.
Take the equations, y* — a^ — Sx* + 9 — 8x = 0,
and y — as — 1=0-
GcHnbining, we will have, for the final equation, 10=0; or^ retaining
the trace of y, 10 + Oy = 0, or y = — y = op.
The given equations are not simultaneous.
1.
2.
GINIAAL XXAMFLBS.
y» — 2ay — 4aj + iB" = 0,
V 4" x— ~4 z^
iln«. 0^ = 4,05^ = 1; y' = 0,y" = 8.
y« — 2xy — 4x — a? = 0y
y« — 2xy — 5« — 2aj» + 2 = 0. _
Am, « = 1, or — 2, y = 1 =fc ^/5, or — 2 drv'— 4.
8.
4.
y-Bi_8y»x* + 8y»«» — V*" + 5y«a — 2y' =
«• — 8yx + 2y« = 0.
Am. X and y indeterminate.
a^_2ya;* + oj" + a5« — 2ya? +y* = 0,
jpt — 2yx + l = 0.
Am. a; = =fcl, ordby^— 8, y = dbl.
606
XLIHINATiaV BXTWIXR TVO ■QirAIIONS
5.
6.
i.
8.
9.
10.
11.
a*
a?
y-
2y»< + SB* + a^ — 2y« + 1=0,
2yx + l=0.
Am, » = {, y=S'
2xy + a^ — 2;y -~ 1 =s 0,
2a!y + a^ + X ^ 0.
Atu. a;&= — 1, <v — I; y = 0, or— |.
2ay + 2a^ — 2y +4 = 0,
4a:y + 7a? — 2j/» — 5 = 0.
il«u. Yaloes imaginaiy.
-4ay + 6a? + 2z + l — 0,
2««>0.
^fu. ass — 1, ysa— 2.
y« + a;y — 2a? + 8a! — 1 = 0,
y»_a? = 0.
Ant. » *■ i, or 1, or }, and y
(ya._l)(a,_2)(y-4)-0,
(a! + 2) (yx — x) (x — 8) m. 0.
Am.
= J, or — l,or — J.
Tallies of x:
VahiMoftr.
«=.— 2
If^-i
a;al
y- + l
x-S
y-+4
z-0
^ inoompadble
a; = 2
y-i
a; incompatible
y inoompatable
a!-.— 2
y-4
x-O
y-4
a;->8
y-4
y«— .2ay + a? + 2ys=0,
^• — 2xy + a^ — 2y — 1 = 0.
jIiw. y = — i^x^ — l drvTE-
lA iVo^— ya^ — 8y^ — 2yx» + 8ya + ley + 2x» — 16 = 0,
^^' I • 1 A
|y»+y-r-fic — 1 a= 0.
Ans. y = ly or — 2 ; a; = {, or — 1.
Jinal eqiiation^ ^ + y — 2 n 0.
or ANT DBORIB. 507
18 |y*-^ + 7«0, (A)
ly" — a? + 8«0. (B)
Ans. X OB 2^ and y ss 1.
After the fint division^ multiply (B) by (p^ — 3)'^ we will get for
the final equation, 9x^ — 14a^ — 27a^ 4- 76 = 0. The only rational
Talne in this equation is x s 2.
14.
y«_aj«_15.0,
3^ — a» — 7-0.
Ans. as = 1| and y — 2.
Final equation, (a^ + lby — (a? + 7)* = 0.
Keqnired the yalnes and the final equation belonging to
IS y' — a^ — 2101 = 0,
^®- y«_aj«_869=.0.
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